# Math Help - Integration Help

1. ## Integration Help

Hey guys, I'm new to integration and these are giving me a really rough time, even just a push in the right direction would be great. Thanks for taking a look! So far, I've only learned the basic integration techniques and u-substitution.

1) $\int\frac{x+1}{\sqrt{x^2+2x-3}}dx$

2) $\int\frac{dx}{x^{\frac{3}{4}}(x^\frac{1}{4}+2)^2}$

I solved 2) and got $16\sqrt[4]{x}-\frac{4}{\sqrt[4]{x}}$. Not sure if this is right, I did expanded completely and brought everything up top before integrating.

3) $\int3x\sin(4-x^2)dx$

4) $\int\frac{dx}{\cos^2x\sqrt{1+\tan{x}}}$

Thanks for any pointers!

2. For the first one, I'd try a $u$-substitution as $u=x^2+2x-3 \Rightarrow du=2x+2 dx$.
Hence the integral is worth $\frac{1}{2} \int \frac{du}{\sqrt u}$. Finish it.
Edit: I also see a u-sub for the third integral : let $u=4-x^2$... can you finish it?

3. Originally Posted by arbolis
For the first one, I'd try a $u$-substitution as $u=x^2+2x-3 \Rightarrow du=2x+2 dx$.
Hence the integral is worth $\frac{1}{2} \int \frac{du}{\sqrt u}$. Finish it.
Edit: I also see a u-sub for the third integral : let $u=4-x^2$... can you finish it?
Wicked I think I worked out a solution for both of them, if you have a sec would you mind checking over my work?

1) $\int\frac{x+1}{\sqrt{x^2+2x-3}}dx$

Let $u=x^2+2x-3$

$du=2x+2dx$

$\frac{1}{2}du=(x+1)dx$

Therefore

$\frac{1}{2}\int\frac{du}{\sqrt{u}}$

$=\frac{1}{2}\int u^{-{\frac{1}{2}}}du$

$=(\frac{1}{2})(2u^{\frac{1}{2}})+c$

$=\sqrt{u}+c$

$\boxed{=\sqrt{x^2+2x-3}+c}$

----------------------------------------------------------
And for number 3) I got the following;
$
\int3x\sin{(4-x^2)}dx$

Let $u=4-x^2$

$du=-2xdx$

$-\frac{3}{2}du=3xdx$

Therefore

$\int(\sin{(u)})(-\frac{3}{2}du)$

$=-\frac{3}{2}\int\sin(u)du$
$
=-\frac{3}{2}(-\cos{u})+C$

$
\boxed{=\frac{3}{2}\cos{(4-x^2)}+C}$

What do you think?

4. Both looks right to me. Very good!
I'll let others helping you for the 2 remaining ones.

5. Cool, thanks!