Question

Evaluate: $\int \left(1-\frac{\log \frac{1}{v}}{v^2}\right)dv$

Answer 1

Consider given the given integration,

$I=\int 1-\frac{\log \frac{1}{v}}{v^2}dv$

$I=\int 1dv-\int \frac{\log \frac{1}{v}}{v^2}dv=v-\int \frac{\log \frac{1}{v}}{v^2}dv$ …..(1)

Let,

$y=\int \frac{\log \frac{1}{v}}{v^2}dv$

Put,

$t=\frac{1}{v}$

$dt=\frac{-1}{v^2}dv$

$dv=-v^{2}dt$

$y=\int \frac{-v^2\log t}{v^2}dv=-\int \left(\log t\right).1dt$

$=-[\log t.t-\int \frac{1}{t}.tdt]$

$=-t.\log t+t+C$

$y=-\frac{1}{v^2}\log \frac{1}{v^2}+\frac{1}{v^2}+C$

Put, the value of in equation (1),we get

$I=v-(-\frac{1}{v^2}\log \frac{1}{v^2}+\frac{1}{v^2}+C)$

$I=v+\frac{1}{v^2}\log \frac{1}{v^2}-\frac{1}{v^2}-C$

Hence, this is the answer,