1. ## Intergal Question

Hi

Can someone tell me where my mistake is for the following questions:

1)Evaluate the integral $\displaystyle \frac{e^{x}}{1+e^{2x}}$
$\displaystyle u = e^{x}$
$\displaystyle u' = e^{x}$
$\displaystyle dx = \frac{du}{e^{x}}$

$\displaystyle \frac{e^{x}}{1+e^{2x}} * \frac{du}{e^{x}}$

$\displaystyle e^{x}$ cancels so i get:

$\displaystyle \frac{1}{1+e^{2x}} + C$
Not sure if this is correct?

2)Evaluate the integral $\displaystyle \sqrt{1 + \sqrt{x}}$
This is what i have done:

$\displaystyle u = \sqrt{x}$
$\displaystyle u' = \frac{1}{2}x^{-0.5}$
$\displaystyle dx = \frac{du}{\frac{1}{2}x^{-0.5}}$

final answer i got was $\displaystyle \frac{4}{3}x^{\frac{3}{2}} + x^2 + C$

P.S

2. for sol.see the attachment

3. Originally Posted by Paymemoney
Hi

Can someone tell me where my mistake is for the following questions:

1)Evaluate the integral $\displaystyle \frac{e^{x}}{1+e^{2x}}$
$\displaystyle u = e^{x}$
$\displaystyle u' = e^{x}$
$\displaystyle dx = \frac{du}{e^{x}}$

$\displaystyle \frac{e^{x}}{1+e^{2x}} * \frac{du}{e^{x}}$

$\displaystyle e^{x}$ cancels so i get:

$\displaystyle \frac{1}{1+e^{2x}} + C$
Not sure if this is correct?
If your going to use substitution, then USE IT. you didn't actually substitute for u in your integral above

$\displaystyle u = e^{x}$ => $\displaystyle du = e^{x}dx$

so integral becomes $\displaystyle \int\frac{u}{1+u^{2}}du$

Now integrate with w.r.t. u and fill back in your substitution for u at the end

4. What does this stand for w.r.t??

5. Originally Posted by Paymemoney
What does this stand for w.r.t??
with respect to.

6. thanks guys i found out the solution.