Integration by u subsitution

Question

So i have this integral

$$\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}x^2 \ln(x^3-3)\,dx.$$

I was thinking of using u subsitution to make everything easier.

I made $u = x^3-3$ and $du = 3x^2dx$.

So I would then re-write my integral as

$$1/3\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}} \ln(x^3-3)\.$$

How would I proceed from here. Should I plug in the integral values? Wouldn't I need to integrate the $\ln()$? Should I use u substitution again? Please help!

I already asked this question, so don't mark it as a duplicate. Unfortunately I couldn't understand what other people were writing.

Answer 1

When you integrate by substitution, you have the express the differential form under the integral sign: $f(x)\,\mathrm dx$ as a differential form $\;g(u)\,\mathrm d u$, and replace the bounds for the integral in $x$ with corresponding bounds for the new variable $u$.

Some details in this case:

If you set $u=x^3-3$, you have $\mathrm d u=3x^2\,\mathrm d x$, so $$\int x^2 \ln(x^3-3)\,dx=\int \ln(x^3-3)(x^2\,dx)=\int\ln u\,\frac13\mathrm du=\frac13\int\ln u\,\mathrm du .$$

Now let's take care of the bounds:

• $x=\sqrt[3]4\leftrightarrow u=4-3=1$,
• $x=\sqrt[3]{3+\mathrm e}\leftrightarrow u=3+\mathrm e-3=\mathrm e$

Thus the integral is $$\frac13\int_1^{\mathrm e}\ln u\,\mathrm du=\frac13(u\ln u-u)\biggr|_1^{\mathrm e}=\frac13.$$

Answer 2

After your substitution, your integral should be $ 1/3 \int_1^e \ln(u) du$.