Question

The value(s ) of $ \int^1_0 \frac { x^4 ( 1 - x )^4 }{ ( 1 + x^2 ) } \, dx $ is are

• A. $ \frac{22}{7} - \pi $
• B. $ \frac {2}{105 } $
• C. 0
• D. $ \frac {71}{ 15} - \frac { 3 \pi }{ 2 } $

Answer 1

The Correct Option is A

$ Let \, \, I = \int^1_0 \frac { x^4 ( 1 - x )^4 }{ ( 1 + x^2 ) } \, dx $
$ = \int^1 _ 0 \frac { ( x^4 - 1 )( 1 - x)^4 + ( 1 - x )^4 }{ ( 1 + x^2 ) } dx $
$ = \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 $
$ = \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 $
$ = \int^1_0 \bigg ( x^6 - 4 \, x^5 + 5 \, x^4 - 4 \, x^2 + 4 - \frac {4}{ 1 + x ^2 } \bigg ) \, dx $
$ = \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$
$ = \frac { 1 } { 7 } - \frac { 4 } { 6 } + \frac { 5 } { 5 } - \frac { 4 } { 3 } + 4 - 4 \bigg ( \frac {\pi}{ 4 } - 0 \bigg ) = \frac {22}{7} - \pi $