# Math Help - More Integrals

1. ## More Integrals

Integral of x^2(3x^3+4)^4

and (x^4)/(3x^5+2)

Thanks! =)

2. Originally Posted by Merdiemae
Integral of x^2(3x^3+4)^4
use substitution. $u = 3x^3 + 4$

and (x^4)/(3x^5+2)
substitution again, $u = 3x^5 + 2$

can you continue? do you see why these substitutions work?

3. No- I know how to find u and du but I'm not sure what to do after that. Do you have to move something outside of the integral sign? I think that's what's throwing me off.

4. Originally Posted by Merdiemae
No- I know how to find u and du but I'm not sure what to do after that. Do you have to move something outside of the integral sign? I think that's what's throwing me off.
here is the first, the second one is done similarly. as for the way we plug in our substitution, some people do it differently. this is the way i like to do it. tell me if you get it

$\int x^2 (3x^3 + 4)^4~dx$

Let $u = 3x^3 + 4$

$\Rightarrow du = 9x^2~dx$

$\Rightarrow \frac 19~du = x^2~dx$

So our integral becomes:

$\frac 19 \int u^4~du$

so as you see. i wrote the 1/9 du instead of x^2 dx, i then replaced what was in the brackets by u, as my substitution demands.

$\frac 19 \int u^4~du = \frac 19 \cdot \frac 15u^5 + C$

$= \frac 1{45}u^5 + C$

Back-substitute:

$= \frac 1{45}(3x^3 + 4)^5 + C$

5. Originally Posted by Merdiemae
No- I know how to find u and du but I'm not sure what to do after that. Do you have to move something outside of the integral sign? I think that's what's throwing me off.
You dont need to do U-substitution neccasarily just realize that through constant manipulation you can get $\int{x^2(3x^3+4)^4dx}=\frac{1}{9}\int{9x^2(3x^3+4) ^4dx}\Rightarrow{\frac{(3x+4)^5}{45}}$

6. Originally Posted by Mathstud28
You dont need to do U-substitution neccasarily just realize that through constant manipulation you can get $\int{x^2(3x^3+4)^4dx}=\frac{1}{9}\int{9x^2(3x^3+4) ^4dx}\Rightarrow{\frac{(3x+4)^5}{45}}$
what you did is what u-substitution imitates. it is more or less the formalization of your method. some students cannot see that however, so substitution is best for beginners i think. but yes, you are correct, that is a valid way to do the problem, and a way Krizalid may have suggested

7. Originally Posted by Jhevon
what you did is what u-substitution imitates. it is more or less the formalization of your method. some students cannot see that however, so substitution is best for beginners i think. but yes, you are correct, that is a valid way to do the problem, and a way Krizalid may have suggested
No...Krizalid didn't suggest it..it just makes sense to me =)...better than U-substitution..but I see what you mean about making it a bit more clear to those new to the subject

8. Thank you very much!! Would the answer to the second one be (1/4)ln(abs(3x^5+2))+c

with u=3x^5+2 and du=4x^3 -> 1/4du=x^3

and 1/4 the integral 1/u du

which would give you (1/4)ln(abs(u))+c

9. Originally Posted by Merdiemae
Thank you very much!! Would the answer to the second one be (1/4)ln(abs(3x^5+2))+c

with u=3x^5+2 and du=4x^3 -> 1/4du=x^3

and 1/4 the integral 1/u du

which would give you (1/4)ln(abs(u))+c
It should be $\frac{1}{15}ln|3x^5+2|+C$

10. Where did the 1/15 come from?

(sorry for asking so many questions)

11. Originally Posted by Mathstud28
It should be $\frac{1}{15}ln|3x^5+2|+C$
I attained this the same way through constant manipulation $\int\frac{x^4}{3x^5+2}dx=\frac{1}{15}\int\frac{15x ^4}{3x^5+2}dx\Rightarrow{\frac{1}{15}ln|3x^5+2|+C}$ I was able to just integrate because once again I had an integral of the form $\int{f(g(x))\cdot{g'(x)}dx}$

12. Originally Posted by Merdiemae
Where did the 1/15 come from?

(sorry for asking so many questions)

13. Originally Posted by Mathstud28
No...Krizalid didn't suggest it..it just makes sense to me =)...better than U-substitution..but I see what you mean about making it a bit more clear to those new to the subject
what i meant was, Krizalid has the taste for these algebraic manipulations to do an integral rather than substitution. he is not very fond of substitution it seems. whether it's trig or u-sub

14. Originally Posted by Jhevon
what i meant was, Krizalid has the taste for these algebraic manipulations to do an integral rather than substitution.

Yeah, I suggest when seein' an obvious chain rule, to modify the integrand and proceed from there, it really takes less time.

Originally Posted by Jhevon
he is not very fond of substitution it seems. whether it's trig or u-sub.
I don't like trig. sub. - I prefer algebraic substitutions, but this is not always possible.