1. ## integration technique

I encounter 2 problems when I explain integration technique to my student.
Could someone help me?
Thx alot.

\int{x^5*[(4-3x^4)]^1/3}=
\int{\frac{e^x}{2+e^(2x)}}=
all with respect to x.

GBU

2. $\int{\frac{e^x}{2 + e^{2x}}\,dx} = \int{\frac{e^x\,dx}{2 + (e^x)^2}}$.

Let $u = e^x$ so that $du= e^x\,dx$ and the integral becomes

$\int{\frac{du}{2 + u^2}}$.

Now you have to use a trigonometric substitution. Let $u = \sqrt{2}\tan{\theta}$ so that $\theta = \arctan{\left(\frac{\sqrt{2}u}{2}\right)}$ and $du = \sqrt{2}\sec^2{\theta}\,d\theta$. The integral becomes

$\int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2 + (\sqrt{2}\tan{\theta})^2}}$

$= \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2 + 2\tan^2{\theta}}}$

$= \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2(1 + \tan^2{\theta})}}$

$= \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2\sec^ 2{\theta}}}$

$= \frac{\sqrt{2}}{2}\int{1\,d\theta}$

$= \frac{\sqrt{2}}{2}\theta + C$

$= \frac{\sqrt{2}}{2}\arctan{\left(\frac{\sqrt{2}u}{2 }\right)} + C$

$= \frac{\sqrt{2}}{2}\arctan{\left(\frac{\sqrt{2}e^x} {2}\right)} + C$.

3. I would use integration by parts on the first integral...

$\int{x^5(4 - 3x^4)^{\frac{1}{3}}\,dx} = -\frac{1}{12}\int{x^2\left(-12x^3\right)(4 - 3x^4)^{\frac{1}{3}}\,dx}$.

Now let $u = x^2$ so that $du = 2x$

Let $dv = -12x^3(4 - 3x^4)^{\frac{1}{3}}$ so that $v = \frac{3}{4}(4 - 3x^4)^{\frac{4}{3}}$ and the integral becomes

$-\frac{1}{12}\int{x^2\left(-12x^3\right)(4 - 3x^4)^{\frac{1}{3}}\,dx} = -\frac{1}{12}\left[\frac{3}{4}x^2(4 - 3x^4)^{\frac{4}{3}} - \int{\frac{3}{2}x(4 - 3x^4)^{\frac{4}{3}}\,dx}\right]$

$= -\frac{1}{16}x^2(4 - 3x^4)^{\frac{4}{3}} + \frac{1}{8}\int{x(4 - 3x^4)^{\frac{4}{3}}\,dx}$.

Now apply integration by parts again.

4. Sir Prove it, I apply the integration by parts again, but I can get the answer. Could u help me end this problem?