Solving the Integral: $\int_{0}^{1}\dfrac{x^4(1-x)^4}{1+x^2}dx$

[Albert1]

\[\int_{0}^{1}\dfrac{x^4(1-x)^4}{1+x^2}dx=?\]

 

[kaliprasad]

\[\int_{0}^{1}\dfrac{x^4(1-x)^4}{1+x^2}dx=?\]

Spoiler

expanding we get

$x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}$

integrating we get

$\frac{x^7}{7} – \frac{2}{3}x^6+ x^5-\frac{4}{3}x^3 + 4x – 4\arctan x$

x=1 gives $\frac{1}{7} – \frac{2}{3}+1-\frac{4}{3} + 4 – \pi = \frac{22}{7}- \pi$ and at x =0 it is zero hence the integral is $\frac{22}{7}- \pi$

 

[Albert1]

Spoiler

expanding we get

$x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}$

integrating we get

$\frac{x^7}{7} – \frac{2}{3}x^6+ x^5-\frac{4}{3}x^2 + 4x – 4\arctan x$

x=1 gives $\frac{1}{7} – \frac{2}{3}+1-\frac{4}{3} + 4 – \pi = \frac{22}{7}- \pi$ and at x =0 it is zero hence the integral is $\frac{22}{7}- \pi$

Spoiler

answer correct , a typo exists in your intergrating

 

[kaliprasad]

Spoiler

answer correct , a typo exists in your intergrating

Spoiler

thanks I have done the needful