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Mathematics Question on Some Properties of Definite Integrals

The value(s ) of ∫01x4(1−x)4(1+x2) dx\int^1_0 \frac { x^4 ( 1 - x )^4 }{ ( 1 + x^2 ) } \, dx is are

A

227−π\frac{22}{7} - \pi

B

2105\frac {2}{105 }

C

0

D

7115−3π2\frac {71}{ 15} - \frac { 3 \pi }{ 2 }

Answer

227−π\frac{22}{7} - \pi

Explanation

Solution

Let  I=∫01x4(1−x)4(1+x2) dxLet \, \, I = \int^1_0 \frac { x^4 ( 1 - x )^4 }{ ( 1 + x^2 ) } \, dx
=∫01(x4−1)(1−x)4+(1−x)4(1+x2)dx= \int^1 _ 0 \frac { ( x^4 - 1 )( 1 - x)^4 + ( 1 - x )^4 }{ ( 1 + x^2 ) } dx
=∫01(x2−1)(1−x)4 dx+∫011+x2+2 x)2(1+x)2= \int^1_0 ( x^2 - 1 )( 1 - x )^4 \, dx + \int^1_0 \frac { 1 + x^2 + 2 \, x )^2 }{ ( 1 + x )^2 }
= \int^1 _ 0 \bigg \\{ ( x^2 - 1 ) ( 1 - x )^4 + ( 1 + x^2 ) - 4 \, x + \frac { 4 \, x^2 }{ ( 1 + x ^2 ) } \bigg \\} \, dx
=∫01((x2−1)(1−x)4+(1+x2)−4 x+4−4 x2(1+x2)) dx= \int^1 _ 0 \bigg ( ( x^2 - 1 ) ( 1 - x )^4 + ( 1 + x^2 ) - 4 \, x + 4 - \frac { 4 \, x^2 }{ ( 1 + x ^2 ) } \bigg ) \, dx
=∫01(x6−4 x5+5 x4−4 x2+4−41+x2) dx= \int^1_0 \bigg ( x^6 - 4 \, x^5 + 5 \, x^4 - 4 \, x^2 + 4 - \frac {4}{ 1 + x ^2 } \bigg ) \, dx
=[x77−4 x66+5 x55−4 x33+4 x−4tan⁡−1 x]01= \bigg [ \frac {x^7}{7} - \frac { 4 \, x^6}{ 6 } + \frac {5 \, x^5 } { 5 } - \frac { 4 \, x^3}{ 3} + 4 \, x - 4 \tan^{-1}\, x \bigg ]^1_0
=17−46+55−43+4−4(π4−0)=227−π= \frac { 1 } { 7 } - \frac { 4 } { 6 } + \frac { 5 } { 5 } - \frac { 4 } { 3 } + 4 - 4 \bigg ( \frac {\pi}{ 4 } - 0 \bigg ) = \frac {22}{7} - \pi