# Math Help - Evaluate the integral (by susbstitution)...

1. ## Evaluate the integral (by susbstitution)...

I have the following integral $\int \frac{-8x}{(x+2)^\frac{1}{4}}$ which Im supposed to evaluate by substitution, but I can't wrap my mind around this,

I tried $u = {(x+2)^{-\frac{1}{4}}}$ but in this case $\frac{du}{dx} = -\frac{1}{4}*{(x+2)^{-\frac{5}{4}}}$, and is getting more complicated...

could you give me a hint, or maybe the substitution is not even possible here?

2. ## Re: Evaluate the integral (by susbstitution)...

Then try the simpler u=x+ 2. du= dx, and x= u- 2. The integral becomes $\int \frac{-8(u- 2)}{u^{1/4}}du= -8\int (u- 2)u^{-1/4}du= -8\int u^{3/4}- 2u^{-1/4} du$

3. ## Re: Evaluate the integral (by susbstitution)...

that's perfect, and the final answer I have got is $\frac{(x+2)^\frac{3}{4}(256-96x)}{21} + C$, I could simplify it a bit, but that changes nothing

thank you!
dokrbb

4. ## Re: Evaluate the integral (by susbstitution)...

Originally Posted by dokrbb
I have the following integral $\int \frac{-8x}{(x+2)^\frac{1}{4}}$ which Im supposed to evaluate by substitution, but I can't wrap my mind around this,

I tried $u = {(x+2)^{-\frac{1}{4}}}$ but in this case $\frac{du}{dx} = -\frac{1}{4}*{(x+2)^{-\frac{5}{4}}}$, and is getting more complicated...

could you give me a hint, or maybe the substitution is not even possible here?
$\text{Here is a better way to do this integral.}$

$\text{Let}~ u=(x+2)^{1/4} ~\Rightarrow~ x=u^4-2 ~\Rightarrow~ dx=4u^3du$

$-8\int{\frac{u^4-2}{u}\cdot 4u^3du}$

$\text{You can pick it up from here.}$

5. ## Re: Evaluate the integral (by susbstitution)...

Originally Posted by thevinh
$\text{Here is a better way to do this integral.}$

$\text{Let}~ u=(x+2)^{1/4} ~\Rightarrow~ x=u^4-2 ~\Rightarrow~ dx=4u^3du$

$-8\int{\frac{u^4-2}{u}\cdot 4u^3du}$

$\text{You can pick it up from here.}$
oh, that's cool also, both are just fine, only a different aproach to the power,

thank you

6. ## Re: Evaluate the integral (by susbstitution)...

Originally Posted by thevinh
$\text{Here is a better way to do this integral.}$

$\text{Let}~ u=(x+2)^{1/4} ~\Rightarrow~ x=u^4-2 ~\Rightarrow~ dx=4u^3du$

$-8\int{\frac{u^4-2}{u}\cdot 4u^3du}$

$\text{You can pick it up from here.}$
I agree that this is an alternative method, but I fail to see why it is better. The substitution is more complicated and so is the differentiation...

7. ## Re: Evaluate the integral (by susbstitution)...

Originally Posted by Prove It
I agree that this is an alternative method, but I fail to see why it is better. The substitution is more complicated and so is the differentiation...
First of all, the above substitution get rid of the fraction power by converting to integer powers which make the integration process simpler. Second, how hard is it to take the derivative of $u^4 - 2$. Finally, integrating $u^6 - 2u^2$ is much simpler than integrating $u^{3/4}-2u^{-1/4}$.