Thread: Integration help

I rewrote this as

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

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

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

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

$\displaystyle u = x+4$

$\displaystyle x = u - 4$

$\displaystyle du = dx$

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

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

So i got this

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

is that right????

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?

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

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.

can anyone check if i got it right.

i got...

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

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.

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