The integral 16_1^2(dx)/(x^3(x^2+2)^2) is equal to:

Question

The integral $16\displaystyle\int_1^2\frac{dx}{x^3(x^2+2)^2}$ is equal to:

Accepted Answer

$I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$

$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$

$=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$

Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$

$=\frac{11}{6}-\ln 4$

Solutions · 1