Evaluate the indefinite integral:
x^2 / (3-4x^3)^6
I got -1/(60(4x^3-3))^5 but it's not right.
Your answer is almost right except that some parentheses are in the wrong places.
Verify the parentheses.
Also, if the question asks to simplify the answer, you may be expected to remove the negative sign by switching the order of the terms in the denominator (to (3-4x³).
Unfortunately computer answers are sometimes quite one-track minded.
The parenthesis are in the wrong spot? I still can't figure it out.
Oh, okay. I got it.
To evaluate the indefinite integral of the function \(\frac{x^2}{{(3-4x^3)}^6}\), you can use the substitution method. Let's go through the steps to find the correct answer.
1. Start by letting \(u = 3 - 4x^3\).
2. Compute the derivative of \(u\) with respect to \(x\): \(\frac{du}{dx} = -12x^2\).
3. Solve the equation from step 1 for \(x\): \(x = \left(\frac{3 - u}{4}\right)^{\frac{1}{3}}\).
4. Substitute \(u\) and \(\frac{du}{dx}\) into the original integral:
\[
\int{\frac{x^2}{{(3-4x^3)}^6}} \, dx = \int{\frac{{\left(\frac{3 - u}{4}\right)^{\frac{2}{3}}}}{{(u)}^6}} \left(-\frac{1}{12}\right)\, du
\]
5. Simplify the integrand:
\[
\int{\frac{{\left(\frac{3 - u}{4}\right)^{\frac{2}{3}}}}{{(u)}^6}} \left(-\frac{1}{12}\right)\, du = -\frac{1}{12} \int{\frac{{(3 - u)^{\frac{2}{3}}}}{{(u)}^6}} \, du
\]
6. Now we can further simplify the integrand by writing \((3 - u)^{\frac{2}{3}}\) as \((u - 3)^{\frac{2}{3}}\):
\[
-\frac{1}{12} \int{\frac{{(u - 3)^{\frac{2}{3}}}}{{(u)}^6}} \, du
\]
7. At this point, we can proceed with partial fraction decomposition or use the binomial expansion. The binomial expansion is generally simpler, so we'll choose that approach.
Start by expanding \((u - 3)^{\frac{2}{3}}\) using the binomial theorem:
\[
(u - 3)^{\frac{2}{3}} = u^{\frac{2}{3}} - 2u^{-\frac{1}{3}} \cdot 3 + 3^2 \cdot u^{-\frac{4}{3}}
\]
8. Now we can rewrite the integral using the expanded form:
\[
-\frac{1}{12} \int{\left(u^{\frac{2}{3}} - 2u^{-\frac{1}{3}} \cdot 3 + 3^2 \cdot u^{-\frac{4}{3}}\right) \frac{1}{{u}^6}} \, du
\]
9. Distribute the fraction across each term:
\[
-\frac{1}{12} \left(\int{\frac{{u^{\frac{2}{3}}}}{{u}^6}} \, du - 6 \int{\frac{{u^{-\frac{1}{3}}}}{{u}^6}} \, du + 9 \int{\frac{{u^{-\frac{4}{3}}}}{{u}^6}} \, du\right)
\]
10. Simplify each integral separately:
\[
-\frac{1}{12} \left(\int{{u}^{-\frac{16}{3}}} \, du - 6 \int{{u}^{-\frac{19}{3}}} \, du + 9 \int{{u}^{-\frac{22}{3}}} \, du\right)
\]
11. Apply the power rule to solve each integral:
\[
-\frac{1}{12} \left(\frac{u^{-\frac{13}{3}}}{-\frac{13}{3}} - \frac{6u^{-\frac{16}{3}}}{-\frac{16}{3}} + \frac{9u^{-\frac{19}{3}}}{-\frac{19}{3}}\right) + C
\]
12. Simplify the expression by multiplying through by the negative exponents:
\[
-\frac{1}{12} \left(\frac{-3u^{-\frac{13}{3}}}{13} + \frac{18u^{-\frac{16}{3}}}{16} - \frac{27u^{-\frac{19}{3}}}{19}\right) + C
\]
13. Replace \(u\) with the original expression for \(x\):
\[
-\frac{1}{12} \left(\frac{-3{(3 - 4x^3)}^{-\frac{13}{3}}}{13} + \frac{18{(3 - 4x^3)}^{-\frac{16}{3}}}{16} - \frac{27{(3 - 4x^3)}^{-\frac{19}{3}}}{19}\right) + C
\]
So, the correct indefinite integral of \(\frac{x^2}{{(3-4x^3)}^6}\) is:
\[
-\frac{1}{12} \left(\frac{-3{(3 - 4x^3)}^{-\frac{13}{3}}}{13} + \frac{18{(3 - 4x^3)}^{-\frac{16}{3}}}{16} - \frac{27{(3 - 4x^3)}^{-\frac{19}{3}}}{19}\right) + C
\]
where \(C\) represents the constant of integration.