{{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^4 \paren {1 - x}^4} {1 + x^2} \rd x       | r = \int_0^1 \dfrac {x^4 \paren {1 - 4 x + 6 x^2 - 4 x^3 + x^4} } {1 + x^2} \rd x       | c = Fourth Power of Difference }} {{eqn | r = \int_0^1 \dfrac {x^4} {1 + x^2} \rd x - 4 \int_0^1 \dfrac {x^5} {1 + x^2} \rd x + 6 \int_0^1 \dfrac {x^6} {1 + x^2} \rd x - 4 \int_0^1 \dfrac {x^7} {1 + x^2} \rd x + \int_0^1 \dfrac {x^8} {1 + x^2} \rd x       | c = Linear Combination of Definite Integrals }} {{end-eqn}} We then establish a reduction formula: {{begin-eqn}} {{eqn | l = \dfrac {x^m} {1 + x^2}       | r = \dfrac {x^{m - 2} \paren {1 + x^2} - x^{m - 2} } {1 + x^2}       | c =  }} {{eqn | n = 1       | r = x^{m - 2} - \dfrac {x^{m - 2} } {1 + x^2}       | c =  }} {{end-eqn}} Then it is a matter of evaluating the individual integrals. {{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^4} {1 + x^2} \rd x       | r = \int_0^1 \paren {x^2 - \dfrac {x^2} {1 + x^2} } \rd x       | c = from $(1)$ }} {{eqn | r = \int_0^1 x^2 \rd x - \int_0^1 \dfrac {x^2} {1 + x^2} \rd x       | c = Linear Combination of Definite Integrals }} {{eqn | r = \intlimits {\dfrac {x^3} 3} 0 1 - \int_0^1 \dfrac {x^2} {1 + x^2} \rd x       | c = Primitive of Power }} {{eqn | r = \intlimits {\dfrac {x^3} 3} 0 1 - \bigintlimits {x - \arctan x} 0 1       | c = Primitive of $\dfrac {x^2} {x^2 + a^2}$ }} {{eqn | r = \dfrac 1 3 - \paren {1 - \arctan 1} - \paren {0 - \arctan 0}       | c = plugging in the limits }} {{eqn | n = 2       | r = \dfrac \pi 4 - \dfrac 2 3       | c = simplifying }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^5} {1 + x^2} \rd x       | r = \int_0^1 \paren {x^3 - \dfrac {x^3} {1 + x^2} } \rd x       | c = from $(1)$ }} {{eqn | r = \int_0^1 x^3 \rd x - \int_0^1 \dfrac {x^3} {1 + x^2} \rd x       | c = Linear Combination of Definite Integrals }} {{eqn | r = \intlimits {\dfrac {x^4} 4} 0 1 - \int_0^1 \dfrac {x^3} {1 + x^2} \rd x       | c = Primitive of Power }} {{eqn | r = \intlimits {\dfrac {x^4} 4} 0 1 - \intlimits {\frac {x^2} 2 - \frac {a^2} 2 \, \map \ln {x^2 + a^2} } 0 1       | c = Primitive of $\dfrac {x^3} {x^2 + a^2}$ }} {{eqn | r = \dfrac 1 4 - \paren {\paren {\dfrac 1 2 - 0} - \paren {\frac 1 2 \, \map \ln {1 + 1^2} - \frac 1 2 \, \map \ln {1 + 0^2} } }       | c = plugging in the limits }} {{eqn | r = \dfrac 1 4 - \dfrac 1 2 + \frac 1 2 \ln 2       | c = simplifying }} {{eqn | n = 3       | r = \dfrac {\ln 2} 2 - \dfrac 1 4       | c = simplifying }} {{eqn | ll= \leadsto       | l = -4 \int_0^1 \dfrac {x^5} {1 + x^2} \rd x       | r = 1 - 2 \ln 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^6} {1 + x^2} \rd x       | r = \int_0^1 \paren {x^4 - \dfrac {x^4} {1 + x^2} } \rd x       | c = from $(1)$ }} {{eqn | r = \int_0^1 x^4 \rd x - \int_0^1 \dfrac {x^4} {1 + x^2} \rd x       | c = Linear Combination of Definite Integrals }} {{eqn | r = \intlimits {\dfrac {x^5} 5} 0 1 - \int_0^1 \dfrac {x^4} {1 + x^2} \rd x       | c = Primitive of Power }} {{eqn | r = \intlimits {\dfrac {x^5} 5} 0 1 - \paren {\dfrac \pi 4 - \dfrac 2 3}       | c = from $(2)$ }} {{eqn | r = \dfrac 1 5 - \dfrac \pi 4 + \dfrac 2 3       | c = plugging in the limits }} {{eqn | n = 4       | r = \dfrac {13} {15} - \dfrac \pi 4       | c = simplifying }} {{eqn | ll= \leadsto       | l = 6 \int_0^1 \dfrac {x^6} {1 + x^2} \rd x       | r = \dfrac {26} 5 - \dfrac {3 \pi} 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^7} {1 + x^2} \rd x       | r = \int_0^1 \paren {x^5 - \dfrac {x^5} {1 + x^2} } \rd x       | c = from $(1)$ }} {{eqn | r = \int_0^1 x^5 \rd x - \int_0^1 \dfrac {x^5} {1 + x^2} \rd x       | c = Linear Combination of Definite Integrals }} {{eqn | r = \intlimits {\dfrac {x^6} 6} 0 1 - \int_0^1 \dfrac {x^5} {1 + x^2} \rd x       | c = Primitive of Power }} {{eqn | r = \intlimits {\dfrac {x^6} 6} 0 1 - \paren {\dfrac {\ln 2} 2 - \dfrac 1 4}       | c = from $(3)$ }} {{eqn | r = \dfrac 1 6 - \dfrac {\ln 2} 2 + \dfrac 1 4       | c = plugging in the limits }} {{eqn | r = \dfrac 5 {12} - \dfrac {\ln 2} 2       | c = simplifying }} {{eqn | ll= \leadsto       | l = -4 \int_0^1 \dfrac {x^7} {1 + x^2} \rd x       | r = 2 \ln 2 - \dfrac 5 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^8} {1 + x^2} \rd x       | r = \int_0^1 \paren {x^6 - \dfrac {x^6} {1 + x^2} } \rd x       | c = from $(1)$ }} {{eqn | r = \int_0^1 x^6 \rd x - \int_0^1 \dfrac {x^6} {1 + x^2} \rd x       | c = Linear Combination of Definite Integrals }} {{eqn | r = \intlimits {\dfrac {x^7} 7} 0 1 - \int_0^1 \dfrac {x^6} {1 + x^2} \rd x       | c = Primitive of Power }} {{eqn | r = \intlimits {\dfrac {x^7} 7} 0 1 - \paren {\dfrac {13} {15} - \dfrac \pi 4}       | c = from $(4)$ }} {{eqn | r = \dfrac 1 7 - \dfrac {13} {15} + \dfrac \pi 4       | c = plugging in the limits }} {{end-eqn}} It remains to gather up the terms. First the terms in $x^5$ and $x^7$, as there is some important cancelling out: {{begin-eqn}} {{eqn | l = \paren {-4 \int_0^1 \dfrac {x^5} {1 + x^2} \rd x} + \paren {-4 \int_0^1 \dfrac {x^7} {1 + x^2} \rd x}       | r = \paren {1 - 2 \ln 2} + \paren {2 \ln 2 - \dfrac 5 3}        | c =  }} {{eqn | r = 1 - \dfrac 5 3       | c =  }} {{eqn | r = -\dfrac 2 3       | c =  }} {{end-eqn}} Thus we continue: {{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {x^4 \paren {1 - x}^4} {1 + x^2} \rd x       | r = \paren {\dfrac \pi 4 - \dfrac 2 3} - \dfrac 2 3 + \paren {\dfrac {26} 5 - \dfrac {3 \pi} 2} + \paren {\dfrac 1 7 - \dfrac {13} {15} + \dfrac \pi 4}       | c =  }} {{eqn | r = \paren {-\dfrac 2 3 - \dfrac 2 3} + \dfrac 1 7 + \dfrac {26} 5 - \dfrac {13} {15} - \pi       | c = rearranging, and resolving the multiples of $\pi$ }} {{eqn | r = \dfrac 1 7 + \dfrac {78 - 13 - 10 - 10} {15} - \pi       | c = gathering all the multiples of $\dfrac 1 {15}$ }} {{eqn | r = 3 + \dfrac 1 7  - \pi       | c = arithmetic }} {{end-eqn}} The result follows. {{qed}}