Values of this integeration

Question

The values of $$I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+x^2}dx$$ My Failed Attempt

I tried applying F(x) =F(a+b-x) in hopes the denominator will cancel out but it's not. I tried to further simplify the expression but with no luck

$$I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+({1-x})^2}dx$$

Adding both integral isn't helping me much

This is one that you just do, rather than look for clever tricks. It has an amusing answer. — Angina Seng, Sep 13 '18 at 16:34
Already asked in 2012, 2015, 2016, 2017. See also Proof that $22/7$ exceeds $\pi$ — Jean-Claude Arbaut, Sep 13 '18 at 17:48

Dr. Sonnhard Graubner · Answer 1 · 2018-09-13T16:32:01.257

2

Use that $$\frac {x^4(1-x)^4}{1+x^2}={x}^{6}-4\,{x}^{5}+5\,{x}^{4}-4\,{x}^{2}+4-4\, \left( {x}^{2}+1 \right) ^{-1} $$

edited Sep 13 '18 at 16:32

answered Sep 13 '18 at 16:26

Just check the edit. The difference between the two integrals is large. – Sarvesh Ravichandran Iyer Sep 13 '18 at 16:28
How did you simply this expression... Did you open up the terms – Aladdin Sep 13 '18 at 16:36
It is the partial fractur decomposition of your integrand – Dr. Sonnhard Graubner Sep 13 '18 at 16:38

score 2 · Accepted Answer · answered Sep 13 '18 at 16:40

$$\int_{0}^1 x^4(1-x)^4\dfrac{1}{1+x^2}$$

First compute the integral without boundaries $$\int x^4(1-x)^4\dfrac{1}{1+x^2}$$

Apply long division on $\dfrac{x^4(1-x)^4}{x^2+1}$ and we get in the form of $x^6-4x^5+5x^4-4x^2+4-\dfrac{4}{x^2+1}$

Now you can easily compute the integral $$\int x^6-4x^5+5x^4-4x^2+4-\dfrac{4}{x^2+1}\ dx$$

2 Answers2