hcju/proofwikiqa huggingface dataset:
Let: {{begin-eqn}} {{eqn | l = z       | r = a^2 - x^2 }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = -2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \implies       | l = \int x \left({\sqrt {a^2 - x^2} }\right)^3 \ \mathrm d x       | r = \int \frac {z^{3/2} } {-2} \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \frac {-z^{5/2} } {\frac 5 2} + C       | c = Primitive of Power }} {{eqn | r = \frac {-z^{5/2} } 5 + C       | c = simplifying }} {{eqn | r = \frac {\left({-\sqrt {a^2 - x^2} }\right)^5} 5 + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \int x^2 \paren {\sqrt {a^2 - x^2} }^3 \rd x       | r = \int \frac {\sqrt z \paren {a^2 - z}^{3/2} } 2 \rd z       | c = Integration by Substitution }} {{eqn | r = \frac {\sqrt z \paren {a^2 - z}^{5/2} } {-6} + \frac {-a^2} {-12} \int \frac {\paren {a^2 - z}^{3/2} } {\sqrt z} \rd z + C       | c = Primitive of $\paren {p x + q}^n \sqrt{a x + b}$ }} {{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {a^2 - x^2}^{3/2} \rd x + C       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 }} {{eqn | o =       | ro= +       | r = \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a} + C       | c = Primitive of $\paren {\sqrt {a^2 - x^2} }^3$ }} {{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {a^2 - x^2} }^3} {24} }} {{eqn | o =       | ro= +       | r = \frac {a^4 x \sqrt {a^2 - x^2} } {16} + \frac {a^6} {16} \arcsin \frac x a + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int x^3 \paren {\sqrt {a^2 - x^2} }^3 \rd x       | r = \int x \paren {x^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x       | c =  }} {{eqn | r = \int x \paren {x^2 - a^2 + a^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x       | c = Primitive of Power }} {{eqn | r = -\int x \paren {a^2 - x^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x       | c = Linear Combination of Integrals }} {{eqn | r = -\int x \paren {\sqrt {a^2 - x^2} }^5 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x       | c = simplifying }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = a^2 - x^2 }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = -2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | o =        | r = -\int x \paren {\sqrt {a^2 - x^2} }^5 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x       | c =  }} {{eqn | r = -\int \frac {\sqrt z z^{5/2} } {-2 \sqrt z} \rd z + a^2 \int \frac {\sqrt z z^{3/2} } {-2 \sqrt z} \rd z       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int z^{5/2} \rd z - \frac {a^2} 2 \int z^{3/2} \rd z       | c = simplifying }} {{eqn | r = \frac 1 2 \frac {z^{7/2} } {7/2} - \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C       | c = Primitive of Power }} {{eqn | r = \frac {z^{7/2} } 7 - a^2 \frac {z^{5/2} } 5 + C       | c = simplifying }} {{eqn | r = \frac {\paren {\sqrt {a^2 - x^2} }^7} 7 - \frac {a^2 \paren {\sqrt {a^2 - x^2} }^5} 5 + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
First: {{begin-eqn}} {{eqn | l = a x^2 + b x + c       | r = \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}       | c = Completing the Square }} {{eqn | n = 1       | ll= \leadsto       | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }       | c =  }} {{end-eqn}} Put: $z = 2 a x + b$ {{begin-eqn}} {{eqn | l = z       | r = 2 a x + b       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 a       | c = Derivative of Power }} {{end-eqn}} Let $D = b^2 - 4 a c$. Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }       | c = from $(1)$ }} {{eqn | r = \int \frac {2 \rd z} {z^2 - D}       | c = Integration by Substitution }} {{end-eqn}} === Negative Discriminant === {{:Primitive of Reciprocal of a x squared plus b x plus c/Negative Discriminant}} {{qed|lemma}} === Positive Discriminant === {{:Primitive of Reciprocal of a x squared plus b x plus c/Positive Discriminant}} {{qed|lemma}} === Zero Discriminant === {{:Primitive of Reciprocal of a x squared plus b x plus c/Zero Discriminant}} {{qed}}
Let: :$a \tan \theta = x$ for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$. From Shape of Tangent Function, this substitution is valid for all real $x$. Then: {{begin-eqn}} {{eqn | l = x       | r = a \tan \theta        | c = from above }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d \theta}       | r = a \sec^2 \theta       | c = Derivative of Tangent Function }} {{eqn | ll= \leadsto       | l = \int \frac 1 {x^2 + a^2} \rd x       | r = \int \frac {a \ \sec^2 \theta} {a^2 \tan^2 \theta + a^2} \rd \theta       | c = Integration by Substitution }} {{eqn | r = \frac a {a^2} \int \frac {\sec^2 \theta} {\tan^2 \theta + 1} rd \theta       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \int \frac {\sec^2 \theta} {\sec^2 \theta} \rd \theta       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 a \int \rd \theta }} {{eqn | r = \frac 1 a \theta + C       | c = Integral of Constant }} {{end-eqn}} As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$: :$\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$ Thus: :$\displaystyle \int \frac 1 {x^2 + a^2} \rd x = \frac 1 a \arctan \frac x a + C$ {{qed|lemma}} When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined. However, consider the limit of the above primitive as $a \to 0$: {{begin-eqn}} {{eqn | l = \lim_{a \mathop \to 0} \frac 1 a \arctan {\frac x a}       | r = \lim_{a \mathop \to 0} \frac {\arctan {\frac x a} } a }} {{eqn | ll= \leadsto       | r = \lim_{a \mathop \to 0} \frac {-x a^{-2} } {1 + \frac {x^2} {a^2} }       | c = L'Hôpital's Rule, Derivative of Arctangent Function }} {{eqn | r = \lim_{a \mathop \to 0} \frac {a^{-2} } {a^{-2} } \frac {-x} {x^2 + a^2} }} {{eqn | r = -\frac 1 x }} {{end-eqn}} This corresponds with the result: :$\displaystyle \int \frac 1 {x^2} \rd x = \frac {-1} x + C$ which follows from Primitive of Power. {{qed}}
We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$. Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$: :$\displaystyle \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$ setting $a := 1, b := 0, c := a^2$: {{begin-eqn}} {{eqn | l = \int \frac 1 {x^2 + a^2} \rd x       | r = \dfrac 2 {\sqrt {4 a^2 - 0} } \map \arctan {\dfrac {2 x + 0} {\sqrt {4 a^2} } } + C       | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$ setting  }} {{eqn | r = \frac 1 a \arctan {\frac x a} + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {\d x} {b x + c}       | c =  }} {{eqn | r = \frac 1 b \ln \size {b x + c} + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{end-eqn}} {{qed}}
Let $b = 0$. From Primitive of Reciprocal of a x squared plus b x plus c, we have: :$\displaystyle \int \frac {\mathrm d x} {a x^2 + b x + c} = \begin{cases} \dfrac 2 {\sqrt {4 a c - b^2} } \arctan \left({\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } }\right) + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \left\vert{\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } }\right\vert + C & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end{cases}$ Let $a c > 0$. Then $b^2 - 4 a c = 0 - 4 a c < 0$ and so: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {a x^2 + 0 x + c}       | r = \frac 2 {\sqrt {4 a c - 0^2} } \arctan \left({\frac {2 a x + 0} {\sqrt {4 a c - 0^2} } }\right) + C       | c =  }} {{eqn | r = \frac 2 {\sqrt {4 a c} } \arctan \left({\frac {2 a x} {\sqrt {4 a c} } }\right) + C       | c =  }} {{eqn | r = \frac 1 {\sqrt {a c} } \arctan \left({x \sqrt {\dfrac a c} }\right) + C       | c = simplifying }} {{end-eqn}} {{qed|lemma}} Let $a c < 0$. Then $b^2 - 4 a c = 0 - 4 a c > 0$ and so: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {a x^2 + 0 x + c}       | r = \frac 1 {\sqrt {0^2 - 4 a c} } \ln \left\vert{\frac {2 a x + 0 - \sqrt {0^2 - 4 a c} } {2 a x + 0 + \sqrt {0^2 - 4 a c} } }\right\vert + C       | c =  }} {{eqn | r = \frac 1 {\sqrt {- 4 a c} } \ln \left\vert{\frac {2 a x - \sqrt {- 4 a c} } {2 a x + \sqrt {- 4 a c} } }\right\vert + C       | c =  }} {{eqn | r = \dfrac 1 {2 \sqrt {-a c} } \ln \left\vert{\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } }\right\vert + C       | c = simplifying }} {{end-eqn}} {{qed|lemma}} Let $c = 0$. Then $b^2 - 4 a c = 0 - 0 = 0$ and so: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {a x^2 + 0 x + 0}       | r = \frac {-2} {2 a x + 0} + C       | c =  }} {{eqn | r = \frac {-1} {a x} + C       | c =  }} {{end-eqn}} {{qed}}
First note that by Derivative of Power: :$(1): \quad \dfrac {\mathrm d} {\mathrm d x} \left({a x^2 + b x + c}\right) = 2 a x + b$ Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {a x^2 + b x + c}       | r = \frac 1 {2a} \int \frac {2 a x \ \mathrm d x} {a x^2 + b x + c}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 {2a} \int \frac {\left({2 a x + b - b}\right) \ \mathrm d x} {a x^2 + b x + c}        | c =  }} {{eqn | r = \frac 1 {2a} \int \frac {\left({2 a x + b}\right) \ \mathrm d x} {a x^2 + b x + c} - \frac b {2a} \int \frac {\mathrm d x} {a x^2 + b x + c}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {2a}\ln \left\vert{a x^2 + b x + c}\right\vert - \frac b {2a} \int \frac {\mathrm d x} {a x^2 + b x + c}       | c = Primitive of Function under its Derivative and $(1)$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {a x^2 + b x + c}       | r = \int \frac 1 a \paren {1 - \frac {b x + c} {a x^2 + b x + c} } \rd x       | c = by division }} {{eqn | r = \frac 1 a \int \rd x - \frac b a \int \frac {x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {\d x} {a x^2 + b x + c}       | c = Linear Combination of Integrals }} {{eqn | r = \frac x a - \frac b a \int \frac {x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {\d x} {a x^2 + b x + c}       | c = Primitive of Constant }} {{eqn | r = \frac x a - \frac b a \paren {\frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c} } - \frac c a \int \frac {\d x} {a x^2 + b x + c}       | c = Primitive of $\dfrac x {a x^2 + b x + c}$ }} {{eqn | r = \frac x a - \frac b {2 a^2} \ln \size {a x^2 + b x + c} - \frac {b^2} {2 a^2} \int \frac {\d x} {a x^2 + b x + c} - \frac {2 a c} {2 a^2} \int \frac {\d x} {a x^2 + b x + c}       | c = multiplying out }} {{eqn | r = \frac x a - \frac b {2 a^2} \ln \size {a x^2 + b x + c} - \frac {b^2 - 2 a c} {2 a^2} \int \frac {\d x} {a x^2 + b x + c}       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^m \rd x} {a x^2 + b x + c}       | r = \int \frac 1 a \frac {a x^m \rd x} {a x^2 + b x + c}       | c =  }} {{eqn | r = \int \frac 1 a \frac {x^{m - 2} a x^2 \rd x} {a x^2 + b x + c}       | c =  }} {{eqn | r = \int \frac 1 a \frac {x^{m - 2} \paren {a x^2 + b x + c - b x - c} \rd x} {a x^2 + b x + c}       | c =  }} {{eqn | r = \frac 1 a \int \frac {x^{m - 2} \paren {a x^2 + b x + c} \rd x} {a x^2 + b x + c} - \frac b a \int \frac {x^{m - 2} x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 a \int x^{m - 2} \rd x - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}       | c = simplification }} {{eqn | r = \frac {x^{m - 1} } {\paren {m - 1} a} - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}       | c = Primitive of Power }} {{end-eqn}} {{qed}}
First: {{begin-eqn}} {{eqn | l = c       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {\d x} {a x^2 + b x}       | c =  }} {{eqn | r = \int \frac {\d x} {x \paren {a x + b} }       | c =  }} {{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C       | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$ }} {{end-eqn}} {{qed}}
Let $c = 0$. From Primitive of Reciprocal of a x squared plus b x plus c, we have: :$\displaystyle \int \frac {\d x} {a x^2 + b x + c} = \begin {cases} \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end{cases}$ As $b \ne 0$ it follows that $b^2 - 4 a c = b^2 > 0$. Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \frac 1 {\sqrt {b^2 - 0} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 0} } {2 a x + b + \sqrt {b^2 - 0} } } + C       | c =  }} {{eqn | r = \frac 1 {\sqrt {b^2} } \ln \size {\frac {2 a x + b - b} {2 a x + b + b} } + C       | c = simplifying }} {{eqn | r = \frac 1 b \ln \size {\frac {a x} {a x + b} } + C       | c = further simplifying }} {{eqn | r = \frac 1 b \paren {\ln \size {\frac x {a x + b} } + \ln \size a} + C       | c = simplifying }} {{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C       | c = subsuming $\dfrac 1 b \ln \size a$ into the arbitrary constant }} {{end-eqn}} {{qed}}
Let $b^2 - 4 a c < 0$. Then: {{begin-eqn}} {{eqn | l = - D       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = - D       | r = q^2       | c = for some $q \in \R$ }} {{eqn | ll= \leadsto       | l = q       | r = \sqrt {4 a c - b^2}       | c = by definition of $D$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {2 \rd z} {z^2 + q^2}       | c = Integration by Substitution }} {{eqn | r = \frac 2 q \arctan \frac z q + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C       | c = substituting for $z$ and $q$ }} {{end-eqn}}
Let $b^2 - 4 a c > 0$. Then: {{begin-eqn}} {{eqn | l = D       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = q^2       | c = for some $q \in \R$ }} {{eqn | ll= \leadsto       | l = q       | r = \sqrt {b^2 - 4 a c}       | c = Definition of $D$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {2 \rd z} {z^2 - q^2}       | c = Integration by Substitution }} {{eqn | r = \frac 1 q \ln \size {\frac {z - q} {z + q} } + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form }} {{eqn | r = \frac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\frac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C       | c = substituting for $z$ and $q$ }} {{end-eqn}}
Let $b^2 - 4 a c = 0$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2}       | c = from $(1)$ }} {{eqn | r = \frac {-4 a} {2 a \paren {2 a x + b} } + C       | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$ }} {{eqn | r = \dfrac {-2} {2 a x + b} + C       | c = simplifying }} {{end-eqn}}
Let: {{begin-eqn}} {{eqn | l = z       | r = 2 a x + b       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a       | c = Derivative of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | r = \int \left({\frac {4 a} {\left({2 a x + b}\right)^2 + 4 a c - b^2} }\right)^2 \ \mathrm d x       | c = Completing the Square }} {{eqn | r = \int \frac {\left({4 a}\right)^2} {\left({z^2 + 4 a c - b^2}\right)^2} \frac {\mathrm d z} {2 a}       | c = Integration by Substitution }} {{eqn | r = 8 a \int \frac {\mathrm d z} {\left({z^2 + 4 a c - b^2}\right)^2}       | c = simplifying }} {{end-eqn}} Let $u = z^2$. Let: {{begin-eqn}} {{eqn | l = u       | r = z^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d z}       | r = 2 z       | c = Derivative of Power }} {{eqn | ll= \implies       | l = 8 a \int \frac {\mathrm d z} {\left({z^2 + 4 a c - b^2}\right)^2}       | r = 8 a \int \frac {\mathrm d u} {2 \sqrt u \left({u + 4 a c - b^2}\right)^2}       | c = Integration by Substitution }} {{eqn | r = 4 a \int \frac {\mathrm d u} {\sqrt u \left({u + 4 a c - b^2}\right)^2}       | c = simplifying }} {{end-eqn}} Recall the result Primitive of $\dfrac 1 {\left({p x + q}\right)^n \sqrt{a x + b} }$: :$\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right)  \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$ Let: {{begin-eqn}} {{eqn | l = x       | o = =:       | r = u }} {{eqn | l = a       | o = =:       | r = 1 }} {{eqn | l = b       | o = =:       | r = 0 }} {{eqn | l = p       | o = =:       | r = 1 }} {{eqn | l = q       | o = =:       | r = 4 a c - b^2 }} {{eqn | l = n       | o = =:       | r = 2 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = 4 a \int \frac {\mathrm d u} {\sqrt u \left({u + 4 a c - b^2}\right)^2}       | r = 4 a \left({\frac {\sqrt u} {\left({4 a c - b^2}\right) \left({u + 4 a c - b^2}\right)} + \frac 1 {2 \left({4 a c - b^2}\right)} \int \frac {\mathrm d u} {\left({u + 4 a c - b^2}\right) \sqrt u} }\right)       | c =  }} {{eqn | r = 4 a \left({\frac z {\left({4 a c - b^2}\right) \left({z^2 + 4 a c - b^2}\right)} + \frac 1 {2 \left({4 a c - b^2}\right)} \int \frac {2 z \ \mathrm d z} {\left({z^2 + 4 a c - b^2}\right) z} }\right)       | c = substituting for $z$ }} {{eqn | r = 4 a \frac {2 a x + b} {\left({4 a c - b^2}\right) \left({\left({2 a x + b}\right)^2 + 4 a c - b^2}\right)} + \frac {2 a} {4 a c - b^2} \int \frac {4 a \ \mathrm d x} {\left({\left({2 a x + b}\right)^2 + 4 a c - b^2}\right)}       | c = substituting for $x$ }} {{eqn | r = \frac {2 a x + b} {\left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)} + \frac {2 a} {4 a c - b^2} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)}       | c = Completing the Square }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = a x^2 + b x + c       | r = \frac {4 a^2 x^2 + 4 a b x + 4 a c} {4 a}       | c = multiplying top and bottom by $4 a$ }} {{eqn | r = \frac {4 a^2 x^2 + 4 a b x + b^2 + 4 a c - b^2} {4 a}       | c =  }} {{eqn | r = \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c       | c =  }} {{eqn | n = 1       | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | o =        | r = \int \frac {x \ \mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {2 a x \ \mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b - b}\right) \ \mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b }\right) \ \mathrm d x} {\left({a x^2 + b x + c}\right)^2} - \frac b {2 a} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b }\right) \ \mathrm d z} {\left({2 a x + b }\right) z^2} - \frac b {2 a} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c = Integration by Substitution from $(1)$ }} {{eqn | r = \frac 1 {2 a} \int \frac {\mathrm d z} {z^2} - \frac b {2 a} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c = simplification }} {{eqn | r = \frac {-1} {2 a z} - \frac b {2 a} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2}       | c = Primitive of Power }} {{eqn | r = \frac {-1} {2 a z} - \frac b {2 a} \left({\frac {2 a x + b} {\left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)} + \frac {2 a} {4 a c - b^2} \int \frac {\mathrm d x} {a x^2 + b x + c} }\right)       | c = Primitive of $\dfrac 1 {\left({a x^2 + b x + c}\right)^2}$ }} {{eqn | r = \frac {-1} {2 a \left({a x^2 + b x + c}\right)} - \frac {b \left({2 a x + b}\right)} {2 a \left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)} - \frac b {4 a c - b^2} \int \frac {\mathrm d x} {a x^2 + b x + c}       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {b^2 - 4 a c - 2 a b x - b^2} {2 a \left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)} - \frac b {4 a c - b^2} \int \frac {\mathrm d x} {a x^2 + b x + c}       | c = common denominator }} {{eqn | r = \frac {- \left({b x + 2 c}\right)} {\left({4 a c - b^2}\right) \left({a x^2 + b x + c}\right)} - \frac b {4 a c - b^2} \int \frac {\mathrm d x} {a x^2 + b x + c}       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | r = \int \frac {c \rd x} {c x \paren {a x^2 + b x + c}^2}       | c = multiplying top and bottom by $c$ }} {{eqn | r = \frac 1 c \int \frac {c \rd x} {x \paren {a x^2 + b x + c}^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 c \int \frac {a x^2 + b x + c - a x^2 - b x} {x \paren {a x^2 + b x + c}^2} \rd x       | c = adding and subtracting $a x^2 + b x$ }} {{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {a x^2 + b x + c}^2} - \frac a c \int \frac {x^2 \rd x} {x \paren {a x^2 + b x + c}^2}       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac b c \int \frac {x \rd x} {x \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | r = \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2}       | c = simplification }} {{eqn | o =        | ro= -       | r = \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | r = - \frac {2 a} {2 c} \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}       | c = splitting up the $\dfrac b c$ term }} {{eqn | o =        | ro= -       | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }       | c =  }} {{eqn | r = - \frac 1 {2 c} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^2}       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }       | c =  }} {{eqn | r = -\frac 1 {2 c} \paren {\frac {-1} {a x^2 + b x + c} }       | c = Primitive of Function under its Derivative }} {{eqn | o =        | ro= -       | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }       | c =  }} {{eqn | r = \frac 1 {2 c \paren {a x^2 + b x + c} }       | c = simplifying }} {{eqn | o =        | ro= -       | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }       | c =  }} {{end-eqn}} {{qed}}