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Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$ $\map P 0$ is true, as from Primitive of $e^{a x}$: :$\displaystyle \int e^{a x} \rd x = \frac {e^{a x} } a$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \int x e^{a x} \rd x       | r = \frac {e^{a x} } a \paren {x - \frac 1 a} + C       | c = Primitive of $x e^{a x}$ }} {{eqn | r = \sum_{k \mathop = 0}^1 \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} }       | c =  }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \int x^r e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } + C$ Then we need to show: :$\displaystyle \int x^{r + 1} e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r + 1 - k} } {a^k} } + C$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \int x^{r + 1} e^{a x} \rd x       | r = \frac {x^{r + 1} e^{a x} } a - \frac {r + 1} a \int x^r e^{a x} \rd x + C       | c = Primitive of $x^n e^{a x}$: Lemma }} {{eqn | r = \frac {x^{r + 1} e^{a x} } a - \frac {r + 1} a \paren {\frac {e^{a x} } a \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } } + C       | c = Induction Hypothesis }} {{eqn | r = \frac {e^{a x} } a \paren {x^{r + 1} - \frac {r + 1} a \paren {\sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } } } + C       | c = factorising $\dfrac {e^{a x} } a$ }} {{eqn | r = \frac {e^{a x} } a \paren {x^{r + 1} - \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {\paren {r + 1} r^{\underline k} x^{r - k} } {a^{k + 1} } } } + C       | c = moving $\dfrac {r + 1} a$ within the summation }} {{eqn | r = \frac {e^{a x} } a \paren {x^{r + 1} - \sum_{k \mathop = 1}^{r + 1} \paren {\paren {-1}^{k - 1} \frac {\paren {r + 1} r^{\underline {k - 1} } x^{r - k + 1} } {a^k} } } + C       | c = changing limits of summation }} {{eqn | r = \frac {e^{a x} } a \paren {x^{r + 1} + \sum_{k \mathop = 1}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r - k + 1} } {a^k} } } + C       | c = {{Defof|Falling Factorial}}, multplying summation by $-1$ }} {{eqn | r = \frac {e^{a x} } a \sum_{k \mathop = 0}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r + 1 - k} } {a^k} } + C       | c = moving initial term into summation }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$ {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = x^n       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = n x^{n - 1}       | c = Derivative of Power }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} } a       | c = Primitive of Exponential of a x }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^n e^{a x} \rd x       | r = x^n \frac {e^{a x} } a - \int \frac {e^{a x} } a \paren {n x^{n - 1} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^n e^{a x} } a - \frac n a \int x^{n - 1} e^{a x} \rd x + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {e^{a x} \rd x} x       | r = \int \frac 1 x \paren {\sum_{k \mathop = 0}^\infty \frac {\paren {a x}^k} {k!} } \rd x       | c = Power Series Expansion for Exponential Function }} {{eqn | r = \int \frac 1 x \paren {1 + \sum_{k \mathop = 1}^\infty \frac {\paren {a x}^k} {k!} } \rd x       | c = extracting the case $k = 0$ from the expansion }} {{eqn | n = 1       | r = \int \frac {\d x} x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \int x^{k - 1} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \ln \size x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \int x^{k - 1} \rd x       | c = Primitive of Reciprocal }} {{eqn | r = \ln \size x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \frac {x^k} k + C       | c = Primitive of Power }} {{eqn | r = \ln \size x + \sum_{k \mathop \ge 1} \frac {\paren {a x}^k} {k \times k!} + C       | c = simplification }} {{end-eqn}} The validity of $(1)$ follows from absolute convergence of the power series expansion. {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = a e^{a x}       | c = Derivative of $e^{a x}$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \frac 1 {x^n}       | c =  }} {{eqn | r = x^{-n}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {x^{-n + 1} } {-n + 1}       | c = Primitive of Power }} {{eqn | r = \frac {-1} {\paren {n - 1} x^{n - 1} }       | c = simplifying }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {e^{a x} \rd x} {x^n}       | r = e^{a x} \paren {\frac {-1} {\paren {n - 1} x^{n - 1} } } - \int \paren {\frac {-1} {\paren {n - 1} x^{n - 1} } } \paren {a e^{a x} } \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {-e^{a x} } {\paren {n - 1} x^{n - 1} } + \frac a {n - 1} \int \frac {e^{a x} \rd x} {x^{n - 1} } + C       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = p + q e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = a q e^{a x}       | c = Derivative of $e^{a x}$ }} {{eqn | r = a \paren {z - p}       | c = in terms of $z$ }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {p + q e^{a x} }       | r = \int \frac {\d z} {a \paren {z - p} z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 a \int \frac {\d z} {z \paren {z - p} }       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \paren {\frac 1 {-p} \ln \size {\frac z {z - p} } } + C       | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$ }} {{eqn | r = \frac {-1} {a p} \ln \size {\frac {p + q e^{a x} } {q e^{a x} } } + C       | c = substituting for $z$ }} {{eqn | r = \frac {-1} {a p} \paren {\ln \size {p + q e^{a x} } - \ln \size {q e^{a x} } } + C       | c = Difference of Logarithms }} {{eqn | r = \frac {-1} {a p} \paren {\ln \size {p + q e^{a x} } - \paren {\ln \size {e^{a x} } + \ln \size q} } + C       | c = Sum of Logarithms }} {{eqn | r = \frac {-1} {a p} \ln \size {p + q e^{a x} } + \frac 1 {a p} \map \ln {e^{a x} } + \frac 1 {a p} \ln \size q + C       | c = $e^{a x}$ always positive }} {{eqn | r = \frac {-1} {a p} \ln \size {p + q e^{a x} } + \frac 1 {a p} \map \ln {e^{a x} } + C       | c = $\dfrac 1 {a p} \ln \size q$ subsumed into arbitrary constant }} {{eqn | r = \frac {-1} {a p} \ln \size {p + q e^{a x} } + \frac 1 {a p} a x + C       | c = Exponential of Natural Logarithm }} {{eqn | r = \frac x p - \frac 1 {a p} \ln \size {p + q e^{a x} } + C       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = p + q e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = a q e^{a x}       | c = Derivative of $e^{a x}$ }} {{eqn | r = a \paren {z - p}       | c = in terms of $z$ }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {p + q e^{a x} }       | r = \int \frac {\d z} {a \paren {z - p} z^2}       | c = Integration by Substitution }} {{eqn | r = \frac 1 a \int \frac {\d z} {z^2 \paren {z - p} }       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \paren {\frac {-1} {-p z} + \frac 1 {\paren {-p}^2} \ln \size {\frac {z - p} z} } + C       | c = Primitive of $\dfrac 1 {x^2 \paren {a x + b} }$ }} {{eqn | r = \frac 1 {a p z} + \frac 1 {a p^2} \ln \size {\frac {z - p} z} + C       | c = simplifying }} {{eqn | r = \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size {\frac {q e^{a x} } {p + q e^{a x} } } + C       | c = substituting for $z$ }} {{eqn | r = \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size {q e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C       | c = Difference of Logarithms }} {{eqn | r = \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size q + \frac 1 {a p^2} \ln \size {e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C       | c = Sum of Logarithms }} {{eqn | r = \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size q + \frac 1 {a p^2} \map \ln {e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C       | c = $e^{a x}$ always positive }} {{eqn | r = \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \map \ln {e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C       | c = $\dfrac 1 {a p^2} \ln \size q$ subsumed into constant }} {{eqn | r = \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} a x - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C       | c = Exponential of Natural Logarithm }} {{eqn | r = \frac x {p^2} + \frac 1 {a p \paren {p + q e^{a x} } } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = a e^{a x}       | c = Derivative of $e^{a x}$ }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {p e^{a x} + q e^{-a x} }       | r = \int \frac 1 {p z + q z^{-1} } \frac {\d z} {a z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 {a p} \int \frac {\d z} {z^2 + \dfrac q p}       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} Let $\dfrac q p > 0$. Then: {{begin-eqn}} {{eqn | l = \frac 1 {a p} \int \frac {\d z} {z^2 + \dfrac q p}       | r = \frac 1 {a p} \paren {\frac 1 {\sqrt {q / p} } \arctan \frac z {\sqrt {q / p} } } + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 1 {a \sqrt {p q} } \map \arctan {\sqrt {\frac p q} z}       | c = simplifying }} {{eqn | r = \frac 1 {a \sqrt {p q} } \map \arctan {\sqrt {\frac p q} e^{a x} }       | c = substituting for $z$ }} {{end-eqn}} Let $\dfrac q p < 0$. Then let $d^2 = -\dfrac q p$ and so: {{begin-eqn}} {{eqn | l = \frac 1 {a p} \int \frac {\d z} {z^2 + \dfrac q p}       | r = \frac 1 {a p} \int \frac {\d z} {z^2 - d^2}       | c =  }} {{eqn | r = \frac 1 {a p} \paren {\frac 1 {2 d} \ln \size {\frac {z - d} {z + d} } } + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$ }} {{eqn | r = \frac 1 {a p} \paren {\frac 1 {2 \sqrt {-q / p} } \ln \size {\frac {z - \sqrt {-q / p} } {z + \sqrt {-q / p} } } } + C       | c = substituting for $d$ }} {{eqn | r = \frac 1 {2 a \sqrt {-p q} } \ln \size {\frac {z - \sqrt {-\frac q p} } {z + \sqrt {-\frac q p} } } + C       | c = simplifying }} {{eqn | r = \frac 1 {2 a \sqrt {-p q} } \ln \size {\frac {e^{a x} - \sqrt {-\frac q p} } {e^{a x} + \sqrt {-\frac q p} } } + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = 1       | c = Derivative of Identity Function }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = e^{a x} \sin b x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2}       | c = Primitive of $e^{a x} \sin b x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x e^{a x} \sin bx \ \mathrm d x       | r = x \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right) - \int \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2}       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac a {a^2 + b^2} \int e^{a x} \sin b x \ \mathrm d x + \frac b {a^2 + b^2} \int e^{a x} \cos b x \ \mathrm d x + C       | c =  }} {{eqn | r = \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2}       | c =  }} {{eqn | o =        | ro= -       | r = \frac a {a^2 + b^2} \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right)       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | o =        | ro= +       | r = \frac b {a^2 + b^2} \left({\frac {e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2} }\right) + C       | c = Primitive of $e^{a x} \cos b x$ }} {{eqn | r = \frac {x e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} - \frac {e^{a x} \left({\left({a^2 - b^2}\right) \sin b x - 2 a b \cos bx}\right)} {\left({a^2 + b^2}\right)^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = 1       | c = Derivative of Identity Function }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = e^{a x} \cos b x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2}       | c = Primitive of $e^{a x} \cos b x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x e^{a x} \sin bx \ \mathrm d x       | r = x \left({\frac {e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2} }\right) - \int \left({\frac {e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2} }\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \frac {x e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2}       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac a {a^2 + b^2} \int e^{a x} \cos b x \ \mathrm d x - \frac b {a^2 + b^2} \int e^{a x} \sin b x \ \mathrm d x + C       | c =  }} {{eqn | r = \frac {x e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2}       | c =  }} {{eqn | o =        | ro= -       | r = \frac a {a^2 + b^2} \left({\frac {e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2} }\right)       | c = Primitive of $e^{a x} \cos b x$ }} {{eqn | o =        | ro= -       | r = \frac b {a^2 + b^2} \left({\frac {e^{a x} \left({a \sin b x - b \cos bx}\right)} {a^2 + b^2} }\right) + C       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | r = \frac {x e^{a x} \left({a \cos b x + b \sin bx}\right)} {a^2 + b^2} - \frac {e^{a x} \left({\left({a^2 - b^2}\right) \cos b x - 2 a b \sin bx}\right)} {\left({a^2 + b^2}\right)^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \ln x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac 1 x       | c = Derivative of Natural Logarithm }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = e^{a x}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \frac {e^{a x} } a       | c = Primitive of $e^{a x}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int e^{a x} \ln x \rd x       | r = \ln x \paren {\frac {e^{a x} } a} - \int \paren {\frac {e^{a x} } a} \frac 1 x \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {e^{a x} \ln x} a - \frac 1 a \int \frac {e^{a x} } x \rd x + C       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} {{qed}}
Note that we have:  {{begin-eqn}} {{eqn	| l = \int_0^1 \ln x \rd x 	| r = \int_0^1 x^0 \paren {\ln x}^1 \rd x }} {{eqn	| r = \frac {\paren {-1}^1 \map \Gamma 2} {1^2} 	| c = Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ }} {{eqn	| r = -1 	| c = Gamma Function Extends Factorial }} {{end-eqn}} We therefore have:  {{begin-eqn}} {{eqn	| l = -1 	| r = \int_0^1 \ln x \rd x }} {{eqn	| r = \frac 1 a \int_0^a \map \ln {\frac u a} \rd u 	| c = substituting $x = \dfrac u a$ where $a > 0$ }} {{eqn	| r = \frac 1 a \int_0^a \paren {\ln u - \ln a} \rd u 	| c = Difference of Logarithms }} {{eqn	| r = \frac 1 a \int_0^a \ln u \rd u - \ln a 	| c = Primitive of Constant, Fundamental Theorem of Calculus }} {{end-eqn}} giving:  :$\displaystyle \frac 1 a \int_0^a \ln u \rd u = \ln a - 1$ so:  :$\displaystyle \int_0^a \ln u \rd u = a \ln a - a$ for all real $a > 0$.  By Fundamental Theorem of Calculus: First Part, we have that:  :$x \ln x - x$ is a primitive of $\ln x$ on $x > 0$. We therefore conclude that, by Primitives which Differ by Constant: :$\displaystyle \int \ln x \rd x = x \ln x - x + C$ for $x > 0$. {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \ln x       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac 1 x       | c = Derivative of $\ln x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^2} 2       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \ln x \rd x       | r = \frac {x^2} 2 \ln x - \int \frac {x^2} 2 \paren {\frac 1 x} \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^2} 2 \ln x - \frac 1 2 \int x \rd x + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 \ln x - \frac 1 2 \paren {\frac {x^2} 2} + C       | c = Primitive of Power }} {{eqn | r = \frac {x^2} 2 \paren {\ln x - \frac 1 2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
From Primitive of $x^m \ln x$: :$\displaystyle \int x^m \ln x \rd x = \frac {x^{m + 1} } {m + 1} \left({\ln x - \frac 1 {m + 1} }\right) + C$ The result follows by setting $m = 1$. {{qed}}
{{begin-eqn}} {{eqn | l = \int x \map \ln {x^2 + a^2} \rd x       | r = \frac {x^2 \map \ln {x^2 + a^2} } 2 - \int \frac {x^3} {x^2 + a^2} \rd x + C       | c = Primitive of $x^m \map \ln {x^2 + a^2}$ with $m = 1$ }} {{eqn | r = \frac {x^2 \map \ln {x^2 + a^2} } 2 - \paren {\frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} } + C       | c = Primitive of $\dfrac {x^3} {x^2 + a^2}$ }} {{eqn | r = \frac {\paren {x^2 + a^2} \map \ln {x^2 + a^2} - x^2} 2 + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \ln x       | c =  }} {{eqn | ll= \implies       | l = \frac {\d u} {\d x}       | r = \frac 1 x       | c = Derivative of $\ln x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = x^m       | c =  }} {{eqn | ll= \implies       | l = v       | r = \frac {x^{m + 1} } {m + 1}       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x^m \ln x \rd x       | r = \frac {x^{m + 1} } {m + 1} \ln x - \int \frac {x^{m + 1} } {m + 1} \left({\frac 1 x}\right) \rd x + C       | c = Integration by Parts }} {{eqn | r = \frac {x^{m + 1} } {m + 1} \ln x - \frac 1 {m + 1} \int x^m \rd x + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^{m + 1} } {m + 1} \ln x - \frac 1 {m + 1} \left({\frac {x^{m + 1} } {m + 1} }\right) + C       | c = Primitive of Power }} {{eqn | r = \frac {x^{m + 1} } {m + 1} \left({\ln x - \frac 1 {m + 1} }\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = \ln x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 x       | c = Derivative of $\ln x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \frac 1 x       | c =  }} {{eqn | ll= \implies       | l = v       | r = \ln x       | c = Primitive of Reciprocal }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\ln x} x \ \mathrm d x       | r = \ln x \ln x - \int \ln x \left({\frac 1 x}\right) \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = \ln^2 x - \int \frac {\ln x} x \ \mathrm d x + C       | c = tidying }} {{eqn | ll= \implies       | l = 2 \int \frac {\ln x} x \ \mathrm d x       | r = \ln^2 x + C       | c = adding $\displaystyle \int \frac {\ln x} x \ \mathrm d x$ to both sides }} {{eqn | ll= \implies       | l = \int \frac {\ln x} x \ \mathrm d x       | r = \frac {\ln^2 x} 2 + C       | c = simplifying }} {{end-eqn}} {{qed}}
From Primitive of $x^m \ln x$: :$\displaystyle \int x^m \ln x \ \mathrm d x = \frac {x^{m + 1} } {m + 1} \left({\ln x - \frac 1 {m + 1} }\right) + C$ Thus: {{begin-eqn}} {{eqn | l = \int \frac {\ln x} {x^2} \ \mathrm d x       | r = \frac {x^{-1} } {-1} \left({\ln x - \frac 1 {-1} }\right) + C       | c = Primitive of $x^m \ln x$, setting $m = -2$ }} {{eqn | r = \frac {-\ln x} x - \frac 1 x + C       | c = simplifying }} {{end-eqn}} {{qed}}
With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = \ln x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = \frac 1 x       | c = Derivative of $\ln x$ }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \ln x       | c =  }} {{eqn | ll= \implies       | l = v       | r = x \ln x - x       | c = Primitive of $\ln x$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \ln^2 x \ \mathrm d x       | r = \int \left({\ln x}\right) \left({\ln x}\right) \ \mathrm d x       | c =  }} {{eqn | r = \left({\ln x}\right) \left({x \ln x - x}\right) - \int \left({x \ln x - x}\right) \frac 1 x \ \mathrm d x + C       | c = Integration by Parts }} {{eqn | r = x \ln^2 x - x \ln x - \int \ln x \ \mathrm d x + \int \ \mathrm d x + C       | c = Linear Combination of Integrals }} {{eqn | r = x \ln^2 x - x \ln x - \left({x \ln x - x}\right) + \int \ \mathrm d x + C       | c = Primitive of $\ln x$ }} {{eqn | r = x \ln^2 x - x \ln x - \left({x \ln x - x}\right) + x + C       | c = Primitive of Constant }} {{eqn | r = x \ln^2 x - 2 x \ln x + 2 x + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = \ln x       | c =  }} {{eqn | l = \frac {\mathrm d z} {\mathrm d x}       | r = \frac 1 x       | c = Derivative of Natural Logarithm }} {{eqn | ll= \implies       | l = \int \frac {\ln^n x} x \ \mathrm d x       | r = \int z^n \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {z^{n + 1} } {n + 1} + C       | c = Primitive of Power }} {{eqn | r = \frac {\ln^{n + 1} x} {n + 1} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = \ln x       | c =  }} {{eqn | l = \frac {\mathrm d z} {\mathrm d x}       | r = \frac 1 x       | c = Derivative of Natural Logarithm }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {x \ln x}       | r = \ln \left({\ln x}\right) + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}