1. ## Integrals! help!

need to get the integral of:

3x - x/((x^2+2)^4/3)

substitution rule here? or is it ln. i'm not sure

2. Hello,

You can integer 3x on its own.

For the second term, you can rewrite it into $- \frac{1}{2} 2x (x^2+2)^{-\frac{4}{3}} = \frac{1}{2} 3* 2x (x^2+2)^{-\frac{4}{3}}$

$-3$ is $\frac{1}{-\frac{4}{3} + 1}$, 2x is the derivate of x²

With all that elements you should be able to find the integral, using only formulas you learnt

3. i dont understand that at all.. i'm new to integrals.. can you show me how to do it directly sorry

4. Originally Posted by s0urgrapes
need to get the integral of:

3x - x/((x^2+2)^4/3)

substitution rule here? or is it ln. i'm not sure
To integrate $\frac{x}{(x^2 + 2)^{4/3}}$, make the substitution $u = x^2 + 2$. Note that $\frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x}$.

So $\int \frac{x}{(x^2 + 2)^{4/3}} \, dx = \int \frac{x}{u^{4/3}} \, \frac{du}{2x} = \frac{1}{2} \, \int \frac{1}{u^{4/3}} \, du = \frac{1}{2} \, \int u^{-4/3} \, du = ....$

Remember to substitute back $u = x^2 + 2$ at the finish ....

5. Ok...

First of all, note the formulas you should know :

The integral of $u'(x) u^n(x)$ is $\frac{1}{n+1} u^{n+1}(x)$

We have here $3x - \frac{x}{(x^2 + 2)^{4/3}} = 3x - \frac{1}{2} 2x (x^2+2)^{-4/3}$

The integral of 3x is 3x²/2

In the second term, u(x)=x²+2. u'(x)=2x.
n=-4/3, so n+1 will be -1/3 and 1/(n+1) will be -3

Hence the integral is :

$\frac{3x^2}{2} - \frac{1}{2} (x^2+2)^{-1/3} * (-3) = \frac{3x^2}{2} + \frac{3}{2} \frac{1}{(x^2+2)^{-1/3}} = \frac{3}{2} (x^2 + \frac{1}{(x^2+2)^{-1/3}})$

6. thanks!