1. Integration help

I rewrote this as

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

2. 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

3. Re: Integration help

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

4. Re: Integration help

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

$u = x+4$

$x = u - 4$

$du = dx$

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

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

5. Re: Integration help

So i got this

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

is that right????

6. Re: Integration help

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?

7. 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

8. Re: Integration help

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.

9. Re: Integration help

can anyone check if i got it right.

i got...

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

10. Re: Integration help

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.

11. 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