# Thread: Stuck in loop using sub method for integrals, work scanned

1. ## Stuck in loop using sub method for integrals, work scanned

I'm doing so much math I think I got a blank on the substitution method for solving integrals.

My work is scanned in the attachment. I think I'm on the right track but there is something I'm forgetting because I'm stuck in a loop.

2. $\displaystyle \int \frac{30x^2}{x^{3}-6} dx$

let $\displaystyle u = x^{3}-6$

then $\displaystyle du = 3x^2 dx \implies dx = \frac{du}{3x^2}$

so you have:

$\displaystyle \int \frac{30x^2}{x^{3}-6} dx= \int \frac{30x^2}{3x^2} \frac{1}{u}du = 10[log(u)]+C = 10[log(x^3-6)]+C$

3. $\displaystyle \displaystyle 5 \cdot \frac{x^5}{x^3-6} =$

$\displaystyle \displaystyle 5 \left( x^2 + \frac{6x^2}{x^3-6}\right) =$

$\displaystyle \displaystyle 5x^2 + 10 \cdot \frac{3x^2}{x^3-6}$

$\displaystyle \displaystyle 5 \int x^2 \, dx + 10\int \frac{3x^2}{x^3-6} \, dx$

$\displaystyle \displaystyle \frac{5x^3}{3} + 10\ln|x^3-6| + C$

4. I can follow these steps and understand them except the part where we get to the last line and we introduce ln. I know 1/u = ln|u| but where does the numerator 3x^2dx go?

Edit: Also, the integral has limits of integration 3 and 4, what do we do with this? Normally when I see limits like this I apply a formula F(b) - F(a), would this be the step after the final step shown above?

5. After supposing $\displaystyle u = x^3-6$, and finding $\displaystyle du = 3x^2 dx$ you get:

$\displaystyle \int \frac{30x^2}{x^{3}-6} dx= \int \frac{30x^2}{3x^2} \frac{1}{u}du$

you should be able to see that the x^2 in the numerator and denominator cancel each other other out, and 30/3 = 10...

you book should have examples on how to use the limits of integration.

$\displaystyle \displaystyle \left[\frac{5x^3}{3} + 10\ln|x^3-6|\right]^4_3$

6. Don't forget (lnf(x))'=F'(x)/F(x).Here:
(3x^2)/(x^3-6)=
(x^3-6)'/(x^3-6)

7. Originally Posted by solidstatemath
I can follow these steps and understand them except the part where we get to the last line and we introduce ln. I know 1/u = ln|u| but where does the numerator 3x^2dx go?

Edit: Also, the integral has limits of integration 3 and 4, what do we do with this? Normally when I see limits like this I apply a formula F(b) - F(a), would this be the step after the final step shown above?
The integral has been done so I won't bother with that. For the limits you may also leave it in terms of u and change the limits using the original subsitution:

$\displaystyle u(3) = 3^3-6 = 21$ and $\displaystyle u(4) = 4^3-6 = 58$

$\displaystyle \left[\: 10 \ln |u| \: \right]^{58}_{21}$

8. Lol yes, the 10 is what is left after simplification... I even did that myself when I was stuck in my loop. :P Thanks.

9. By using limits of integration I apply F(4) - F(3) and this is what I get:

10ln58 - 10ln21 = ~10.16 and it's wrong.

What could I be doing wrong?

10. Originally Posted by solidstatemath
By using limits of integration I apply F(4) - F(3) and this is what I get:

10ln58 - 10ln21 = ~10.16 and it's wrong.

What could I be doing wrong?
you forgot the term $\displaystyle \displaystyle \frac{5x^3}{3}$ in the antiderivative