Question

express the integrand as a sum of partial fractions and evaluate the integrals.
int\frac{x + 4}{2x^{3}-8x}dx
rewrite the integrand as the sum of partial fractions.
\frac{x + 4}{2x^{3}-8x}=square

Explanation:

Step1: Factor the denominator

Factor $2x^{3}-8x = 2x(x^{2} - 4)=2x(x - 2)(x + 2)$.
Then assume $\frac{x + 4}{2x^{3}-8x}=\frac{A}{2x}+\frac{B}{x - 2}+\frac{C}{x + 2}$.

Step2: Clear the fractions

Multiply both sides by $2x^{3}-8x$ to get $x + 4=A(x - 2)(x + 2)+B(2x)(x + 2)+C(2x)(x - 2)$.

Step3: Find the value of A

Let $x = 0$, then $0+4=A(0 - 2)(0 + 2)+B(0)+C(0)$, so $4=A(- 4)$, and $A=-1$.

Step4: Find the value of B

Let $x = 2$, then $2 + 4=A(0)+B(2\times2)(2 + 2)+C(0)$, so $6 = B(16)$, and $B=\frac{3}{8}$.

Step5: Find the value of C

Let $x=-2$, then $-2 + 4=A(0)+B(0)+C(2\times(-2))(-2 - 2)$, so $2 = C(16)$, and $C=\frac{1}{8}$.
So $\frac{x + 4}{2x^{3}-8x}=-\frac{1}{2x}+\frac{3}{8(x - 2)}+\frac{1}{8(x + 2)}$.

Answer:

$-\frac{1}{2x}+\frac{3}{8(x - 2)}+\frac{1}{8(x + 2)}$