- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 3,084

\int_0^4 {\frac{\sqrt{t}}{t+1}}dt

$

$\displaystyle

U=\sqrt{t}\ \ \ t=U^2 \ \ \ dt=2Udu

$

$\displaystyle

\frac{\sqrt{t}}{t+1} \Rightarrow \frac{U}{U^2+1}

$

$\displaystyle

\int_0^4 \frac{U}{U^2+1} 2Udu

$

if ok so far tried $U= sec^2(\theta)$

but couldn't not get answer which is