1. ## Integrations by parts

Intergrate these:

(x^2) dx
(2+3x)^1/2

and

(x^2-25)^1/2 dx
x

We have to intergrate them using parts and it's so frustrating because i just don't know how to do it! please help, thank you!

2. $\displaystyle \int \frac{x^2}{\sqrt{3x + 2}} ~dx = \int x^2 ~(3x + 2)^{-\frac{1}{2}} ~dx$.

Remember the rule for integration by parts: $\displaystyle \int u~dv = u~v - \int v~du$

Letting $\displaystyle u = x^2$ and $\displaystyle dv = (3x + 2)^{-\frac{1}{2}}$, so $\displaystyle du = 2x$ and $\displaystyle v = \frac{2}{3} \sqrt{3x + 2}$

$\displaystyle = \frac{2}{3}x^2~\sqrt{3x + 2} - \int \frac{4}{3}x~\sqrt{3x + 2}~dx$

Now we will leave the first part of this equation alone just for now and deal with the second part which requires another integration by parts.

Let $\displaystyle u = x$ and $\displaystyle v = \sqrt{3x + 2}$. So $\displaystyle du = 1$ and $\displaystyle v = \frac{2}{9} (3x + 2)^{\frac{3}{2}}$

$\displaystyle \frac{4}{3}~\int x~\sqrt{3x + 2}~dx = \frac{4x}{3} \times\frac{2}{9} (3x + 2)^{\frac{3}{2}} - \frac{4}{3}\int~\frac{2}{9} (3x + 2)^{\frac{3}{2}} ~dx$

$\displaystyle = \frac{8x}{27} (3x + 2)^{\frac{3}{2}} - \frac{8}{27} \times \frac{(3x + 2)^{\frac{5}{2}}}{\frac{5}{2} \times 3}$

$\displaystyle = \frac{8x}{27} (3x + 2)^{\frac{3}{2}} - \frac{16}{405} (3x + 2)^{\frac{5}{2}}$

Now, returning to the equation we left off from, the solution to this problem is:

$\displaystyle = \frac{2}{3}x^2~\sqrt{3x + 2} - \int \frac{4}{3}x~\sqrt{3x + 2}~dx$

$\displaystyle = \frac{2}{3}x^2~\sqrt{3x + 2} - \frac{8x}{27} (3x + 2)^{\frac{3}{2}} + \frac{16}{405} (3x + 2)^{\frac{5}{2}}$

If needed it can be simplified:

$\displaystyle = \frac{2}{3}\sqrt{3x + 2}~\left[x^2 - \frac{4x}{9} (3x + 2) + \frac{8}{135} (3x + 2)^2\right]$

$\displaystyle = \frac{2}{405}\sqrt{3x + 2}~\left[135x^2 - 60x (3x + 2) + 8 (3x + 2)^2\right]$

$\displaystyle = \frac{2}{405}\sqrt{3x + 2}~(27x^2 - 24x + 32)$

Now you try do the other one...

3. thanks i get it now!