Evaluate $\int (x^2-6)^3 (2x)\,dx$ using $u$ substitution

Question

Section 5.2

Can somebody verify my solution? Thanks

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Evaluate $\int (x^2-6)^3 (2x)dx$ using $u$ substitution

Let $u = x^2-6$. Then $\frac{du}{dx} = 2x$

and so, by multiplying both sides of the equation by $dx$ and then dividing by $2x$:

$\frac{du}{2x} = dx$

$\rightarrow \int (x^2-6)^3(2x)dx = \int u^3 2x \frac{du}{2x} = \int u^3 du$

And finally,

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

Answer 1

It is correct. A very useful thing you can do with indefinite integrals is to derive (when possible if the calculations aren't too much) the result, if it gives you the integrand function then your result is correct for sure. Indeed $$\frac{\text{d}}{\text{d}x}\left[\frac{(x^2-6)^4}{4}+C\right]=4\frac{(x^2-6)^3}{4}(2x)=(x^2-6)^3(2x)$$

Answer 2

Yes, this is correct (and revert the substitution in the end!). Please consider using an online-calculator (like WolframAlpha or SymbolLab) for verifying your solutions instead of asking multiple question of the same kind within a short period of time (see a relevant post here on MSE.Meta).

Note that SymbolLab offers a step-by-step solution aswell.