An Integral.......

Let's first simplify x^4[1 - x]^4 / [1 + x^2]

[1 - x]^4  =  [x - 1]^4  =  x^4 - 4x^3 + 6x^2 - 4x + 1

So .....  x^4[ 1 - x]^4 / [ 1 + x^2]   =  x^4 [ x^4 - 4x^3 + 6x^2 - 4x + 1] / [1 + x^2] =

[ x^8 - 4x^7 + 6x^6 - 4x^5 + x^4] / [ 1 + x^2]

Perform synthetic division

                x^6  - 4x^5  + 5x^4 - 4x^2 + 4

x^2 + 1  [ x^8- 4x^7 + 6x^6 - 4x^5 + x^4]

               x^8                x^6      

              _________________________

                      -4x^7  + 5x^6 - 4x^5  + x^4

                     -4x^7               -4x^5

                    _______________________

                                  5x^6               + x^4

                                  5x^6               + 5x^4

                                 __________________

                                                         -4x^4

                                                         -4x^4   - 4x^2

                                                        ___________

                                                                       4x^2   

                                                                       4x^2 + 4

                                                                      _______

                                                                              -4

So......we have

1

∫     x^6  - 4x^5  + 5x^4 - 4x^2 + 4 - 4 / [ x^2 + 1]   dx      = 

0

          1                      1             1                   1          1                       1

x^7/7 ]      -  (2/3)x^6 ]    +  x^5 ]     - (4/3)x^3 ]  + 4x ]    -  4arctan(x) ]           =

         0                        0             0                  0         0                        0

(1/7)  -         (2/3)        +     1           -   (4/3)        +    4     -   4arctan(1)   =

22/7     -  4  [pi / 4]  =

22/7  - pi    ≈  0.0013

Integrate: ∫x^4[1 - x]^4 / [1 + x^2]dx,from x=0 to 1

\(\displaystyle\int_0^1\;\frac{x^4(1-x)^4}{1-x^2}\;dx\\ =\displaystyle\int_0^1\;\frac{x^4(1-x)^4}{(1-x)(1+x)}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^4(x-1)^3}{x+1}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^4(x^3-3x^2+3x-1)}{x+1}\;dx\\ =-\displaystyle\int_0^1\;\frac{x^7-3x^6+3x^5-x^4}{x+1}\;dx\\ \text{Do the algebraic division}\\ =-\displaystyle\int_0^1\;x^6-4x^5+7x^4-6x^3+6x^2-6x+6-\frac{6}{x+1}\;dx\\ \)

\( =-\displaystyle\int_0^1\;x^6-4x^5+7x^4-6x^3+6x^2-6x+6-\frac{6}{x+1}\;dx\\ =-\left[\; \frac{x^7}{7}-\frac{4x^6}{6}+\frac{7x^5}{5}-\frac{6x^4}{4}+\frac{6x^3}{3}-\frac{6x^2}{2}+6x-6ln(x+1)\;\right]_0^1\\ =-\left[\; \frac{x^7}{7}-\frac{2x^6}{3}+\frac{7x^5}{5}-\frac{3x^4}{2}+2x^3-3x^2+6x-6ln(x+1)\;\right]_0^1\\ =-\left[\; \frac{1}{7}-\frac{2}{3}+\frac{7}{5}-\frac{3}{2}+2-3+6-6ln(2)\;\right]\\ =-\left[\; \frac{1}{7}-\frac{2}{3}+\frac{7}{5}-\frac{3}{2}+5-6ln(2)\;\right]\)

1/7-2/3+7/5-3/2+5 = 4.3761904761904762 = 919/210

\(=6ln2-\frac{919}{210}\)

-4.3761904761904762+6*log(2) = -2.570010502206589

This is not correct, I have made some silly mistake somewhere, but the method is correct :)

Wolfram Alpha says the answer is approx 0.0023

Mmmm.....WA  seems to get the same result that I have :

https://www.wolframalpha.com/input/?i=x%5E4%5B1+-+x%5D%5E4+%2F+%5B1+%2B+x%5E2%5Ddx,from+x%3D0+to+1

Compute the definite integral:
 integral_0^1 ((1 - x)^4 x^4)/(x^2 + 1) dx
For the integrand ((1 - x)^4 x^4)/(x^2 + 1), cancel common terms in the numerator and denominator:
 = integral_0^1 ((x - 1)^4 x^4)/(x^2 + 1) dx
For the integrand ((x - 1)^4 x^4)/(x^2 + 1), do long division:
 = integral_0^1 (x^6 - 4 x^5 + 5 x^4 - 4 x^2 - 4/(x^2 + 1) + 4) dx
Integrate the sum term by term and factor out constants:
 = -4 integral_0^1 1/(x^2 + 1) dx + integral_0^1 x^6 dx - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of 1/(x^2 + 1) is tan^(-1)(x):
 = (-4 tan^(-1)(x)) right bracketing bar _0^1 + integral_0^1 x^6 dx - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-4 tan^(-1)(x)) right bracketing bar _0^1 = (-4 tan^(-1)(1)) - (-4 tan^(-1)(0)) = -π:
 = -π + integral_0^1 x^6 dx - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^6 is x^7/7:
 = -π + x^7/7 right bracketing bar _0^1 - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 x^7/7 right bracketing bar _0^1 = 1^7/7 - 0^7/7 = 1/7:
 = 1/7 - π - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^5 is x^6/6:
 = 1/7 - π + (-(2 x^6)/3) right bracketing bar _0^1 + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-(2 x^6)/3) right bracketing bar _0^1 = (-(2 1^6)/3) - (-(2 0^6)/3) = -2/3:
 = -11/21 - π + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^4 is x^5/5:
 = -11/21 - π + x^5 right bracketing bar _0^1 - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 x^5 right bracketing bar _0^1 = 1^5 - 0^5 = 1:
 = 10/21 - π - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^2 is x^3/3:
 = 10/21 - π + (-(4 x^3)/3) right bracketing bar _0^1 + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-(4 x^3)/3) right bracketing bar _0^1 = (-(4 1^3)/3) - (-(4 0^3)/3) = -4/3:
 = -6/7 - π + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of 1 is x:
 = -6/7 - π + 4 x right bracketing bar _0^1
Evaluate the antiderivative at the limits and subtract.
 4 x right bracketing bar _0^1 = 4 1 - 4 0 = 4:
Answer: |= 22/7 - π