Integration help

**asilvester635** · Dec 11th 2012, 09:54 AM

I rewrote this as

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

**Zouzz** · Dec 11th 2012, 10:02 AM

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

**asilvester635** · Dec 11th 2012, 10:41 AM

yah it should be ^-1/2 but i made a mistake on putting ^4

**skeeter** · Dec 11th 2012, 10:43 AM

$\int \frac{2x+1}{\sqrt{x+4}} \, dx$

$u = x+4$

$x = u - 4$

$du = dx$

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

$\int 2u^{1/2} - 7u^{-1/2} \, du$

**asilvester635** · Dec 11th 2012, 10:46 AM

So i got this

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

is that right????

**skeeter** · Dec 11th 2012, 10:48 AM

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?

**asilvester635** · Dec 11th 2012, 11:29 AM

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

**Prove It** · Dec 11th 2012, 02:46 PM

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.

**asilvester635** · Dec 13th 2012, 03:12 PM

can anyone check if i got it right.

i got...

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

**skeeter** · Dec 13th 2012, 03:17 PM

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.

**x3bnm** · Dec 17th 2012, 09:37 PM

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

Thread: Integration help

LinkBack

Thread Tools

Display

Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Re: Integration help

Similar Math Help Forum Discussions

Integration help please these are three small even basic integration problems

sketch the region of integration and change the order of integration.

Definite Integration - Double integration by parts

Simple Integration problem: Trigonometric Integration

Tricky integration, possibly Integration Factor

Search Tags