Anyone want to help me with this integral, I am 95 % sure I need to use a combination of u-substitution / integration by parts, I just get stuck with what to do the u-sub on...

Question

owikbenstan · Answer

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unklerhaukus · Answer

$$\int e^{2x}\sqrt{e^x-1}\ \mathrm dx\
=\int e^x\sqrt{e^x-1}e^x \ \mathrm dx$$
let $u =e^x-1$
   $\mathrm du = e^x\ \mathrm dx$
$$=\int (u+1)\sqrt{u}\ \mathrm du\
=$$

owikbenstan · Answer

Wait, you put $$e^x -1$$ where it should be $$e^x +1$$

unklerhaukus · Answer

oh right,

unklerhaukus · Answer

$$\int e^{2x}\sqrt{e^x+1}\ \mathrm dx\
=\int e^x\sqrt{e^x+1}e^x \ \mathrm dx$$
let $u=e^x+1$
   $\mathrm du=e^x\ \mathrm dx$
$$=\int (u-1)\sqrt{u}\ \mathrm du\
=$$

unklerhaukus · Answer

(sorry about that)

unklerhaukus · Answer

can you solve from here?

owikbenstan · Answer

Pretty sure yeah, I had the u-sub correct, however I did not solve for $$e^x$$ , I guess it didn't click because it seems like a function in it's own right, not a variable..

mww · Answer

you generally use a substitution of a difficult expression (in this case bit inside square root)

owikbenstan · Answer

I solved the expression, I originally wanted to use IBP on the last step only to realize that I could simply distribute and use the power rule, thanks for the help guys!