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

Question

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.

https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80#Details_of_evaluation_of_the_integral — Winther, Jan 07 '17 at 14:44
This seems to be rather popular integral: Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$, Shortest method for $\int^{1}_{0}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$, Creative way for $\int _0 ^1 \frac{x^4 (1-x)^4} {x^2 +1} {\rm d}x$. Found using Approach0. — Martin Sleziak, Jan 07 '17 at 15:15

Bernard · Answer 1 · 2017-01-05T10:40:24.333

4

Hint:

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

edited Jan 05 '17 at 10:40

answered Jan 05 '17 at 10:35

Bernard

1

Sir it is $(1-x)^4$. – Jan 05 '17 at 10:37
Oops! I misread the formula. Fixed. Thanks for pointing it. – Bernard Jan 05 '17 at 10:41

score 4 · Answer 2 · answered Jan 05 '17 at 10:38

4

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.

answered Jan 05 '17 at 10:38

sTertooy

score 3 · Accepted Answer · answered Jan 05 '17 at 10:51

3

Hint -

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

answered Jan 05 '17 at 10:51

Kanwaljit Singh

3 Answers3