Question

Integration by Partial Fractions

1.) ∫4dx/(x^4-1)

2.) ∫8dx/x(x^2+2)^2

Answer 1

1) Solution:

$\frac{4}{x^4-1} = \frac{4}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^2+1} = \newline \frac{A(x+1)(x^2+1) + B(x-1)(x^2+1) + C(x-1)(x+1)}{(x-1)(x+1)(x^2+1)} = \newline \frac{A(x^3+x^2+x+1) + B(x^3-x^2+x-1) + C(x^2-1)}{(x-1)(x+1)(x^2+1)} = \newline \frac{x^3(A+B)+x^2(A-B+C)+x(A+B)+(A-B-C)}{(x-1)(x+1)(x^2+1)}$

from nominator

$A+B=0 \newline A-B+C=0 \newline A+B=0 \newline A-B-C=4$

solution of this system is: A=1 B=-1 C=-2, so

$\frac{4}{x^4-1} = \frac{1}{x-1} - \frac{1}{x+1} - \frac{2}{x^2+1}$

and

$\int \frac{4}{x^4-1} dx = \int \frac{1}{x-1} dx - \int \frac{1}{x+1} dx - 2 \int \frac{2}{x^2+1} dx = \newline \ln |x-1| - \ln |x+1| - 2 \arctan x + C$

2) Solution:

$\int \frac{8}{x(x^2+2)^2} dx = \int \frac{8x}{x^2(x^2+2)^2} dx = \int \frac{4}{x^2(x^2+2)^2} d(x^2)$

Using substitution $s=x^2$ and denoting s to x for convenience we will receive: $\int \frac{4}{x(x+2)^2} dx$

$\frac{4}{x(x+2)^2} = \frac{A}{x} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2 + Bx(x+2) + Cx}{x(x+2)^2} = \frac{x^2(A + B) + x(4A + 2B + C) + 4A}{x(x+2)^2}$

from nominator

$A+B=0 \newline 4A + 2B + C=0 \newline 4A=8$

solution of this system is: A=2 B=-2 C=-4, so

$\frac{4}{x(x+2)^2} = \frac{2}{x} - \frac{2}{x+2} - \frac{4}{(x+2)^2}$

and

$\int \frac{4}{x(x+2)^2} dx = \int \frac{2}{x} dx - \int \frac{2}{x+2} dx - \int \frac{4}{(x+2)^2} dx = 2 \ln |x| - 2 \ln |x+2| + \frac{4}{x+2} + C$