How do you Integrate #(x^2-1)*(x^4+4x)^6dx# using u-substitution?

Question

Isabella Ackerman · Accepted Answer

I don't believe you do.  Check the question again.

Samantha Adkison · Answer

undefined To integrate the expression $ \int (x^2-1)(x^4+4x)^6 \, dx $ using u-substitution, let $ u = x^4 + 4x $. Then, $ du = (4x^3 + 4) \, dx $. Rearrange to solve for $ dx $ to get $ dx = \frac{du}{4(x^3 + 1)} $. Substitute these into the integral to get:

$$ \int (x^2-1)(x^4+4x)^6 \, dx = \int (x^2 - 1)u^6 \cdot \frac{du}{4(x^3 + 1)} $$

This simplifies to:

$$ \frac{1}{4} \int \frac{(x^2 - 1)u^6}{x^3 + 1} \, du $$

Now, we need to split the fraction into two separate fractions:

$$ \frac{1}{4} \int \left(\frac{x^2u^6}{x^3 + 1} - \frac{u^6}{x^3 + 1}\right) \, du $$

Next, integrate each term separately:

$$ \frac{1}{4} \left(\int \frac{x^2u^6}{x^3 + 1} \, du - \int \frac{u^6}{x^3 + 1} \, du\right) $$

For the first integral, we use another substitution: $ v = x^3 + 1 $, $ dv = 3x^2 \, dx $, $ \frac{1}{3} dv = x^2 \, dx $. This gives us:

$$ \frac{1}{4} \left(\frac{1}{3} \int \frac{u^6}{v} \, dv - \int \frac{u^6}{v} \, dv\right) $$

$$ = \frac{1}{4} \left(\frac{1}{3} \ln|v| - \ln|v| \right) + C $$

$$ = \frac{1}{4} \left(\frac{1}{3} - 1\right) \ln|v| + C $$

$$ = \frac{-1}{6} \ln|v| + C $$

$$ = \frac{-1}{6} \ln|x^3 + 1| + C $$

Thus, the integral becomes:

$$ \frac{-1}{6} \ln|x^3 + 1| + \frac{1}{4} \int \frac{u^6}{x^3 + 1} \, du + C $$

For the second integral, it's a straightforward integral:

$$ \frac{1}{4} \int \frac{u^6}{x^3 + 1} \, du = \frac{1}{4} \int \frac{u^6}{v} \, dv $$

$$ = \frac{1}{4} \ln|v| + C $$

$$ = \frac{1}{4} \ln|x^3 + 1| + C $$

Now, putting it all together, the final result is:

$$ \frac{-1}{6} \ln|x^3 + 1| + \frac{1}{4} \ln|x^3 + 1| + C $$

$$ = \frac{1}{12} \ln|x^3 + 1| + C $$

Where $ C $ is the constant of integration.