# Thread: Help with integral using ln

1. ## Help with integral using ln

Well, i was assigned to do this problem but i haven't yet covered the ln formula and my professor still wants me to do it.
Can someone please show me how to solve this particular integration (See Attachment)?

Thank you.

2. $\int\frac{1}{\sqrt{x+\sqrt{x}}}dx$

$let ..x=u^4...dx=4u^3$

$\int\frac{1}{\sqrt{(\sqrt{x}\sqrt{x}+\sqrt{x}}}dx$

$\int\frac{1}{\sqrt{(\sqrt{x}(\sqrt{x}+1))}}dx$

$\int\frac{1}{x^{\frac{1}{4}}\sqrt{(\sqrt{x}+1)}}dx$

now use the substitute

$\int\frac{4u^3}{u\sqrt{(u^2+1)}}du$

$\int\frac{4(u^2)}{\sqrt{(u^2+1)}}du$

$now..let...tany=u... sec^2y dy=du...secy=\sqrt{1+u^2}$

$4u^2=4(u^2+1-1)=4(sec^y-1)=4tan^2u...$

3. Thank you so very much

4. One last thing.

So is the final answer (4(tan^2)y)/secy ?

5. Originally Posted by Amer
[tex]

$\int\frac{4u^3}{u\sqrt{(u^2+1)}}du$

$\int\frac{4(u^2)}{\sqrt{(u^2+1)}}du$

$now..let...tany=u... {\color{red}{sec^2y dy=du}}...secy=\sqrt{1+u^2}$

$4u^2=4(u^2+1-1)=4(sec^2y-1)=4tan^2u...$
$\int\left(\frac{4tan^2y}{secy}\right)(sec^2y)dy$

6. So after that I solve it like a normal integration or is that how it stays?

7. no you can solve it

$\int\left(\frac{4tan^2y}{secy}\right)(sec^2y)dy$

$\int4tan^2y(secy)dy$ ok

$4\int(sec^2y-1)(secy)dy$

$4\int(sec^3y-secy)dy$

$4\left(\int sec^3ydy - \int secydy\right)$

do you want me to continue or you can do that...

8. ok

$\int sec^3y dy - \int secy dy$

first multiply the denominator and numerator with sec(y)+tan(y)

$\int secy\left(\frac{secy+tany}{secy+tany}\right) dy$

$\int \frac{sec^2y+secy(tany)}{secy+tany} dy$

now you have the numerator is the derive of the denominator so the integral is ln(denominator) since the integrate of f'(x)/f(x) is ln(f(x)

this is the first
now

$\int sec^3y dy$

let dv=sec^2y and u=secy
v=tany.....du=secy tany

$\int sec^3y dy=secy(tany)-\int secy(tan2^y)dy$

$\int sec^3y dy=secy(tany)-\int secy(sec^2y-1)dy$

$\int sec^3y dy=secy(tany)-\left(\int sec^3y dy-\int secy dy)\right)$

$\int sec^3y dy=secy(tany)-\int sec^3y dy+\int secy dy$

$2\int sec^3y dy = secy(tany)+\int secy dy$

$\int sec^3y dy=\frac{secy(tany)+\int secy dy}{2}$

and the integral in the right we integrate it before