# How to evaluate the integral using integration by parts with the indicated choices of u and dv. int (lny)/sqrt(y) dy ?

Refer to explanation

#### Explanation:

Using integration by parts we have

int lny/sqrtydy=int 2*lny*(sqrty)'dy=2lny*sqrty-2int sqrty*(lny)'dy= 2lnysqrty-2int sqrty*(1/y)dy=2lnysqrty-2int(1/sqrty)=2lnysqrty-4*int (sqrty)'dy=2lnysqrty-4sqrty=2sqrty(lny-2)

Hence finally we get

$\int \ln \frac{y}{\sqrt{y}} \mathrm{dy} = 2 \sqrt{y} \cdot \left(\ln y - 2\right)$

Remarks
1. For two functions f(x),g(x) integration by parts is

$\int f \left(x\right) \cdot \left(g \left(x\right)\right) ' \mathrm{dx} = f \left(x\right) \cdot g \left(x\right) - \int g \left(x\right) \cdot \left(f \left(x\right)\right) ' \mathrm{dx}$

1. The phrase $\left(\right) '$ donates first derivative