# Thread: solving an integral using the table

1. ## solving an integral using the table

Hi,

Could someone please tell me if this is correct?

int. e^2x/sq.2+e^x

int. du/a^2+b^2=1/a arctanu/a+c
u=2+e^x
du=e^xdx

int. e^2x/sq. 2+e^x=1/sq.2 arctan 2+e^x/sq. 2+c

Thank you

2. It's about $\displaystyle\int\frac{e^{2x}}{2^2+e^x}dx$ or $\displaystyle\int\frac{e^{2x}}{\sqrt{2+e^x}}dx$ ?

3. I think it's the second one.

Just set $u=\sqrt{2+e^x}$

4. This one

Sorry for the confusion

5. Just set
Do I just plug this into the forumula or do I need to find du?

Is du 1/2(2+e^x)^-1/2???

Thank you

6. Originally Posted by chocolatelover
Is du 1/2(2+e^x)^-1/2???
You forgot the chain rule, try again

7. Is du 1/2(2+e^x)^-1/2???
You forgot the chain rule
OOPs!

so du is 1/2(2+e^x)^-1/2(e^x). This won't work for u substitution, right? It wouldn't just be 1/sq. 2art sq. 2+e^x/sq. 2+e^x+c, would it?

Thank you

8. $\int\frac{e^{2x}}{\sqrt{2+e^x}}\,dx$

If $u=\sqrt{2+e^x}\implies du=\frac{e^x}{2\sqrt{2+e^x}}\,dx$, the integral becomes to

$\int\frac{e^{2x}}{\sqrt{2+e^x}}\,dx=2\int(u^2-2)\,du$

That's easy, don't you?

9. $u=\sqrt{2+e^x}\Rightarrow u^2=2+e^x\Rightarrow 2udu=e^xdx$
Substitute and you'll get $\displaystyle2\int(u^2-2)du$

10. Originally Posted by chocolatelover
Hi,

Could someone please tell me if this is correct?

int. e^2x/sq.2+e^x

int. du/a^2+b^2=1/a arctanu/a+c
u=2+e^x
du=e^xdx

int. e^2x/sq. 2+e^x=1/sq.2 arctan 2+e^x/sq. 2+c

Thank you
Please do us a favor and either learn LaTeX or start using parenthesis! The way you are writing this is confusing.

-Dan