1. ## Help Solving Integral

Before I ask the problem, I also want to know how to post or what program to use to show the integral sign and square root sign etc...

Now to the problem

X^7 / fifth root of 2X^4+5

If possible, please show using the U-Substitution Method. If another is easier, post as well.

2. Originally Posted by gammaman
Before I ask the problem, I also want to know how to post or what program to use to show the integral sign and square root sign etc...

Now to the problem

X^7 / third root of 2X^4+5

If possible, please show using the U-Substitution Method. If another is easier, post as well.
To check to make sure what you're talking about, you have
x^7 / (2x^4 +5)^1/3 dx ?
If so, start by substituting
x^4 for u
x^4 = u
4x^3dx = du
dx = du/4x^3

Substitute back in. We're only going to substitute dx at first
x^7/ (2x^4 +5)^1/3 * du/4x^3
Note that the x^7/4x^3 will reduce to x^4/4, which allows us to substitute it for u.

1/4 (integral ( u/(2u +5)^1/3 du)) = 1/4 * 3/40 (2u +5)^(2/3) * (4u -15)

Substitute u back in and we arrive at

1/4 * 3/40 * (2x^4 +5)^(2/3) * (4x^4 -15)

Hope that helped. -Edit-

3. Originally Posted by gammaman
Before I ask the problem, I also want to know how to post or what program to use to show the integral sign and square root sign etc...

[snip]

4. Originally Posted by gammaman

Now to the problem

X^7 / fifth root of 2X^4+5
$\int{\frac{x^{7}}{\sqrt[5]{2x^{4}+5}}\,dx}=\int{\frac{x^{4}\cdot x^{3}}{\sqrt[5]{2x^{4}+5}}\,dx},$ and then put $u^5=2x^4+5.$

5. Originally Posted by gammaman
Before I ask the problem, I also want to know how to post or what program to use to show the integral sign and square root sign etc...

Now to the problem

X^7 / fifth root of 2X^4+5

If possible, please show using the U-Substitution Method. If another is easier, post as well.
When you edit a question after getting a reply you make the person who gave the reply look foolish. You originally posted third root .....

6. Originally Posted by mr fantastic
When you edit a question after getting a reply you make the person who gave the reply look foolish. You originally posted third root .....

Yes I realized that but it was too late, sorry.

7. That's not a problem actually, it's always a worth of try, may be that'd help some other people.

But, the idea is essentially the same for both integrals. Both methods work.

8. Originally Posted by Krizalid
$\int{\frac{x^{7}}{\sqrt[5]{2x^{4}+5}}\,dx}=\int{\frac{x^{4}\cdot x^{3}}{\sqrt[5]{2x^{4}+5}}\,dx},$ and then put $u^5=2x^4+5.$
would you mind showing me how to solve it through?

9. $5u^4\,du=8x^3\,dx$ and $\frac58u^4\,du=x^3\,dx.$ Furthermore, we have $x^4=\frac{u^5-5}{2}$ and $\sqrt[5]{2x^4+5}$ becomes $u.$ Now put these together.