# Evaluating $\int_0^1 \frac{x^4 (1-x)^4}{1+x^2} dx$

We have to solve the following $$\int_0^1 \frac{x^4 (1-x)^4}{1+x^2} dx$$

I tried to substitute $x =\tan m$, but in that I got stuck.

Hint -

Simplify term by multiplying $x^4(1-x)^2(1-x)^2$ then divide by $1+x^2$. Then easily integrate it.

Apply long division on the integrand to obtain $$\int_0^1 \left(x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1+x^2}\right)dx,$$ which is easy to solve.

• reduces to $\frac{22}{7}-\pi.$ – Maverick Jan 5 '17 at 13:26

Hint:

No substitution required here: just divide $x^4(1-x)^4$ by $x^2+1$.

• Sir it is $(1-x)^4$. – Rohan Jan 5 '17 at 10:37
• Oops! I misread the formula. Fixed. Thanks for pointing it. – Bernard Jan 5 '17 at 10:41