Evaluate using any method Number 2

[Nawz]

Homework Statement

Evaluate using any method:

$$\int\frac{8x^3+10} {\sqrt[3]{5x-4}}$$dx

Homework Equations

The Attempt at a Solution

I'm lost on this one.

 

[HallsofIvy]

Since it is that cube root in the denominator that is causing the trouble, a kind of obvious substitution is $u= 5x- 4$. Then du= 5dx or dx= (1/5)du.

Also, 5x= u+ 4 so x= (u+ 4)/5. Replacing x by that in $8x^3+ 10$ gives you a cubic, in u, in the numerator divided by $\sqrt[3]{u}= u^{1/3}$. That reduces to a sum of terms involving $u^{3- 1/3}= u^{8/3}$, $u^{2- 1/3}= u^{5/3}$, and $u^{1- 1/3}= u^{-2/3}$.

 

[Nawz]

Where do you get the 5x=u+4?

 

[Mark44]

From the substitution u = 5x - 4.

 

[Nawz]

So do you get something like:

8$$\int\frac{(u+4/5)^3 + 10}{u^(1/3)}$$

That's U raised to the 1/3 on the denominator
?

 

[Mark44]

It's not u + 4/5 in the numerator -- it's (1/5)(u + 4). Also, where is du in your integral? You seem to be ignoring it.

 

[Nawz]

It's not u + 4/5 in the numerator -- it's (1/5)(u + 4). Also, where is du in your integral? You seem to be ignoring it.

Yea that's what I meant. I don't understand where the Du goes.

8$$\int\frac{((u+4)/5)^3 + 10}{u^(1/3)}$$

So that is right? Then does du takes the spot of the 5?

 

[hunt_mat]

Use u=5x-4 then du=5dx, insert this into the integral to obtain:
$$\int\frac{8x^3+10} {\sqrt[3]{5x-4}}dx=\frac{1}{5}\int\frac{8((u+4)/5)^{3}+10}{u^{1/3}}du$$
Now all you have to do is expand the cube and integrate using
$$\int x^{n}dx=\frac{1}{n+1}x^{n+1}+c$$

 

[Nawz]

What is n?

 

[hunt_mat]

$$n$$ is a number, like 3 or 2/3.