Answer:
[tex]\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c[/tex]
Step-by-step explanation:
[tex]\int\frac{x^{4}}{x^{4} -1}dx[/tex]
Adding and Subtracting 1 to the Numerator
[tex]\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx[/tex]
Dividing Numerator seperately by [tex]x^{4} - 1[/tex]
[tex]\int 1 + \frac{1}{x^{4}-1 }\, dx[/tex]
Here integral of 1 is x +c1 (where c1 is constant of integration
[tex]x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx[/tex]----------------------------------(1)
We apply method of partial fractions to perform the integral
[tex]\frac{1}{(x-1)(x+1)(x^{2}+1)}[/tex] = [tex]\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}[/tex]------------------------------------------(2)
[tex]\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}[/tex]
1 = [tex]A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)[/tex]-------------------------(3)
Substitute x= 1 , -1 , i in equation (3)
1 = A(1+1)(1+1)
A = [tex]\frac{1}{4}[/tex]
1 = B(-1-1)(1+1)
B = [tex]-\frac{1}{4}[/tex]
1 = C(i-1)(i+1)
C = [tex]-\frac{1}{2}[/tex]
Substituting A, B, C in equation (2)
[tex]\int\frac{x^{4}}{x^{4} -1}dx[/tex] = [tex]\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }[/tex]
On integration
Here [tex]\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx[/tex]
[tex]\int\frac{x^{4}}{x^{4} -1}dx[/tex] = [tex]\frac{1}{4} ln(x-1)[/tex] - [tex]\frac{1}{4} ln(x+1)[/tex] - [tex]\frac{1}{2} arctanx[/tex] + c2---------------------------------------(4)
Substitute equation (4) back in equation (1) we get
[tex]x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2[/tex]
Here c1 + c2 can be added to another and written as c
Therefore,
[tex]\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c[/tex]
