# Integration help

Printable View

• Dec 11th 2012, 09:54 AM
asilvester635
Integration 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 AM
Zouzz
Re: Integration help
You need to rewrite them as $\int (2x+1)\cdot (x+4)^{-\frac{1}{2}}\text{ dx}$ and use integration by parts, so choose $u(x)=(x+4)^{-\frac{1}{2}}$ and $v(x)=2x+1$, then

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

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

Feel free to ask, if anything I just wrote need explanation :)
• Dec 11th 2012, 10:41 AM
asilvester635
Re: Integration help
yah it should be ^-1/2 but i made a mistake on putting ^4
• Dec 11th 2012, 10:43 AM
skeeter
Re: Integration help
$\int \frac{2x+1}{\sqrt{x+4}} \, dx$

$u = x+4$

$x = u - 4$

$du = dx$

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

$\int 2u^{1/2} - 7u^{-1/2} \, du$
• Dec 11th 2012, 10:46 AM
asilvester635
Re: Integration help
So i got this

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

is that right????
• Dec 11th 2012, 10:48 AM
skeeter
Re: Integration help
Quote:

Originally Posted by asilvester635
So i got this

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

is that right????

check antiderivatives by taking the derivative ... do you get back to the integrand?
• Dec 11th 2012, 11:29 AM
asilvester635
Re: 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 PM
Prove It
Re: Integration help
Quote:

Originally Posted by asilvester635
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

Integration by Parts is different to Substitution. Substitution is used as the opposite of the Chain Rule for derivatives, while Integration by Parts is used as the opposite of the Product Rule for derivatives.
• Dec 13th 2012, 03:12 PM
asilvester635
Re: 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 PM
skeeter
Re: Integration help
Quote:

Originally Posted by asilvester635
can anyone check if i got it right.

i got...

= 2(x + 4)^1/2 - 7(x + 4)^-1/2 + c

check it yourself ... take the derivative of your solution.
• Dec 17th 2012, 09:37 PM
x3bnm
Re: Integration help
We know that $\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx$

More compactly:

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

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

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

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

\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*}

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

Check:

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