I rewrote this as

= (2x + 1)(x + 1)^4 is that right??? then i just combine them and integrate after??????????????????

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- Dec 11th 2012, 09:54 AMasilvester635Integration help
I rewrote this as

= (2x + 1)(x + 1)^4 is that right??? then i just combine them and integrate after?????????????????? - Dec 11th 2012, 10:02 AMZouzzRe: Integration help
You need to rewrite them as $\displaystyle \int (2x+1)\cdot (x+4)^{-\frac{1}{2}}\text{ dx}$ and use integration by parts, so choose $\displaystyle u(x)=(x+4)^{-\frac{1}{2}}$ and $\displaystyle v(x)=2x+1$, then

$\displaystyle \int u(x)\cdot v(x) \text{ dx}=v(x)\cdot U(x)-\int v'(x)\cdot U(x)\text{ dx}$,

where $\displaystyle v'(x)$ is the differential of $\displaystyle v(x)$ and $\displaystyle U(x)$ is the antiderivative of $\displaystyle u(x)$.

Feel free to ask, if anything I just wrote need explanation :) - Dec 11th 2012, 10:41 AMasilvester635Re: Integration help
yah it should be ^-1/2 but i made a mistake on putting ^4

- Dec 11th 2012, 10:43 AMskeeterRe: Integration help
$\displaystyle \int \frac{2x+1}{\sqrt{x+4}} \, dx$

$\displaystyle u = x+4$

$\displaystyle x = u - 4$

$\displaystyle du = dx$

$\displaystyle \int \frac{2(u-4) + 1}{\sqrt{u}} \, du$

$\displaystyle \int 2u^{1/2} - 7u^{-1/2} \, du$ - Dec 11th 2012, 10:46 AMasilvester635Re: Integration help
So i got this

= (2)2(x + 4)^1/2

is that right???? - Dec 11th 2012, 10:48 AMskeeterRe: Integration help
- Dec 11th 2012, 11:29 AMasilvester635Re: Integration help
oh i get it!!!!! yes integration by substitution..... i wasn't familiar with the word integration by parts that's why... THANK YOU SO MUCH

- Dec 11th 2012, 02:46 PMProve ItRe: Integration help
- Dec 13th 2012, 03:12 PMasilvester635Re: Integration help
can anyone check if i got it right.

i got...

= 2(x + 4)^1/2 - 7(x + 4)^-1/2 + c - Dec 13th 2012, 03:17 PMskeeterRe: Integration help
- Dec 17th 2012, 09:37 PMx3bnmRe: Integration help
We know that $\displaystyle \int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$

More compactly:

$\displaystyle \int u \, dv=uv-\int v \, du.\!$

For $\displaystyle \int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx $ Let $\displaystyle dv=\frac{1}{\sqrt{x+4}}$ and $\displaystyle u=2x+1$

$\displaystyle \therefore v = 2\sqrt{x+4}\;\; \text{ and } du = 2\;\;dx$

where $\displaystyle v$ is the antiderivative of $\displaystyle dv$ and $\displaystyle du$ is the derivative of $\displaystyle u$

$\displaystyle \begin{align*}\int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx =& (2x+1)2\sqrt{x+4} - \int 4\sqrt{x+4} \;\;dx\\ =& 2(2x+1)\sqrt{x+4} -\frac{8}{3}(x+4)^{\frac{3}{2}}+C\\ =& \frac{2}{3}\sqrt{x+4}(2x-13)+C.......\text{[After simplification]}\end{align*}$

$\displaystyle \therefore\int \frac{(2x+1)}{\sqrt{x+4}}\;\;dx = \frac{2}{3}\sqrt{x+4}(2x-13)+ C$

Check:

http://www.wolframalpha.com/input/?i=integration+(2*x%2B1)(x%2B4)^(-1%2F2)+dx