Question

evaluate the integral, using integration by parts. $int\frac{ln x}{x^{4}}dx$

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \ln x$ and $\mathrm{d}v=\frac{1}{x^{4}}\mathrm{d}x$. Then $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$ and $v=-\frac{1}{3x^{3}}$ (since $\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n+1}+C,n
eq - 1$, here $n=-4$).

Step2: Substitute into the formula

$\int\frac{\ln x}{x^{4}}\mathrm{d}x=uv-\int v\mathrm{d}u=-\frac{\ln x}{3x^{3}}-\int(-\frac{1}{3x^{3}})\cdot\frac{1}{x}\mathrm{d}x$.

Step3: Simplify the remaining integral

$\int(-\frac{1}{3x^{3}})\cdot\frac{1}{x}\mathrm{d}x=-\frac{1}{3}\int x^{-4}\mathrm{d}x$. Using the power - rule for integration $\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n + 1}+C$, we have $-\frac{1}{3}\int x^{-4}\mathrm{d}x=-\frac{1}{3}\cdot\frac{x^{-4 + 1}}{-4+1}=\frac{1}{9x^{3}}+C$.

Answer:

$-\frac{\ln x}{3x^{3}}-\frac{1}{9x^{3}}+C$