What is the Integral of Square Root of T over T+1 with Substitution?

[karush]

$\displaystyle
\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

 

[Krizalid1]

When you perform a substitution with a definite integral, you need to change the bounds. So for $0\le t\le4$ you get $0\le u\le 2$ since $u=\sqrt t.$ Now, note that $\dfrac{{{u}^{2}}}{{{u}^{2}}+1}=\dfrac{{{u}^{2}}+1-1}{{{u}^{2}}+1}=1-\dfrac{1}{{{u}^{2}}+1},$ so if you integrate the right side, you'll see immediate integrals there.

 

[karush]

When you perform a substitution with a definite integral, you need to change the bounds. So for $0\le t\le4$ you get $0\le u\le 2$ since $u=\sqrt t.$ Now, note that $\dfrac{{{u}^{2}}}{{{u}^{2}}+1}=\dfrac{{{u}^{2}}+1-1}{{{u}^{2}}+1}=1-\dfrac{1}{{{u}^{2}}+1},$ so if you integrate the right side, you'll see immediate integrals there.

So this will go to
$2\left[U-\tan^{-1}(U)\right]_0^4$
$2\left[√t-\tan^{-1}√t\right]_0^4=1.7857$

 

[MarkFL]

You have the correct result, but you could have just used:

\(\displaystyle 2\left[u-\tan^{-1}(u) \right]_0^2=2\left(2-\tan^{-1}(2) \right)\approx1.78570256441182\)

Once you make a substitution in a definite integral and have rewritten the integrand, differential and limits in terms of the new variable, there is no need to consider the old variable again. :D

 

[karush]

I that about that...
Still need more practice I get stuck easy with these...

Mahalo