Answer
a) See the explanation.
b) $\int \frac{\ln x}{x^2}dv=-\frac{\ln x}{x}-\frac{1}{x}+C$
Work Step by Step
Part a)
Here is the Quotient Rule:
$\frac{d(\frac{u}{v})}{dx}=\frac{\frac{du}{dx}\cdot v-u\cdot \frac{dv}{dx}}{v^2}$
Equivalently,
$\frac{d(\frac{u}{v})}{dx}=\frac{1}{v}\frac{du}{dx}-\frac{u}{v^2}\frac{dv}{dx}$
Multiplying both sides with $dx$,
$d(\frac{u}{v})=\frac{1}{v}du-\frac{u}{v^2}dv$
Integrating both sides,
$\int d(\frac{u}{v})=\int \frac{1}{v}du-\int \frac{u}{v^2}dv$
$\frac{u}{v}=\int \frac{1}{v}du-\int \frac{u}{v^2}dv$
Thus,
$\int \frac{u}{v^2}dv=-\frac{u}{v}+\int \frac{1}{v}du$
Part b)
Take $u=\ln x$ and $v=x$.
Then,
$\int \frac{\ln x}{x^2}dv=-\frac{\ln x}{x}+\int \frac{1}{x}\cdot \frac{1}{x}dx$
$\int \frac{\ln x}{x^2}dv=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx$
$\int \frac{\ln x}{x^2}dv=-\frac{\ln x}{x}-\frac{1}{x}+C$
