Question

Integrate x^4/x^4-16

Answer 1

Here it is Hope it helps you..

Answer 2

$\bold{\int{\frac{x^4}{x^4-16}}\,dx} \\\\ = \bold{\int{1+\frac{16}{x^4-16}}\,dx}$
$= \bold{\int{1}\,dx}+\bold{16\int{\frac{1}{x^4-16}\,dx}}$
$=\bold{x+16\int{\frac{1}{x^4-16}\,dx}}$
now, we have to integrate , 1/(x⁴ - 16)
first break in smallest factors of it
e.g., (x⁴ - 16) = (x -2)(x + 2)(x² + 4)
now, use partial fraction concept ,
e.g.,
1/(x⁴ - 16) = A/(x - 2) + B/(x + 2) + (Cx + D)/(x² + 4)
Multiply (x⁴ - 16) both sides,
1 = A(x + 2)(x² + 4) + B(x -2)(x² + 4) + (Cx + D)(x² - 4)
1 = (x² + 4)(Ax + 2A + Bx - 2B) + (Cx + D)(x² - 4)
1 = 2(A - B)x² + 8(A - B) + (A +B)x³ + 4(A + B)x + Cx³ - 4Cx + Dx² - 4D
1 = (A + B + C)x³ + (2A -2B + D)x² + 2(A + B -D)x + 4(2A - 2B - D)

now, compare it ,
A + B + C = 0-----(1)
2A+ D = 2B-------(2)
(A + B - D) = 0 -----(3)
8A - 8B - 4D = 1 -----(4)
after solving it , we get ,
A = 1/32 , B = -1/32 , C = 0, D = -1/8

now,
$\bold{\int{\frac{1}{x^4-16}}\,dx=\int{\frac{A}{x-2}}\,dx+\int{\frac{B}{x+2}}\,dx+\int{\frac{Cx+D}{x^2+4}}\,dx}$
$\bold{=\int{\frac{1}{32(x-2)}}\,dx+\int{\frac{1}{ - 32(x+2)}}\,dx+\int{\frac{1}{ - 8(x^2+4)}}\,dx}$
$=\bold{\frac{ln(x-2)}{32}-\frac{ln(x+2)}{32}-\int{\frac{1}{8(x^2+4)}}\,dx}$

we know,
$\bold{\int{\frac{1}{x^2+a^2}}\,dx=\frac{1}{2a}tan^{-1}\frac{x}{a}+C}$
use this ,
then,
$=-\bold{\frac{1}{8}\int{\frac{1}{x^2+4}}\,dx=-\frac{1}{16}tan^{-1}\frac{x}{2}}$


hence, answer is
$\int1/(x^4 - 16)dx = 1/32ln|x - 2| - 1/32ln|x + 2| - 1/16tan^{ - 1} (x/2) + C$