Need help solving an integral with substitution method

Question

I'm currently working on a calculus problem and I'm having trouble solving it. The integral I'm trying to evaluate is:

$$\int^{}_{} \left( 2x+3\right)^{2} \ln \left( 2x+3\right) dx$$

I know that I can use the substitution method to simplify this integral. Letting $u=2x+3$, I can rewrite the integral as:

$$\frac{1}{2} \int (u)^2 \ln u, du$$

From here, I can use integration by parts to solve the integral. However, I'm not sure about how to proceed after this point.

Could someone please provide guidance on how to complete this integral using the substitution method? Any help would be greatly appreciated. Thank you in advance!

Answer 1

$I=\int u^2\ ln(u)\ du=\frac1{3}\int ln(u)\ du^3=\frac 1{3}[u^3\ ln(u)-\int u^3\ d(ln(u))]$

$I=\frac 1{3}[u^3\ ln(u)-\int u^2\ du]=\frac 1{3}[u^3\ ln(u)-\frac 1{3} u^3]+C$

Answer 2

Ok so resuming where you left off now we can integrate this using the power rule and integration by parts. Let $v=\log(u)$, then $dv=1/u \cdot du$, and let $u=u^2/2$, then $du=u \cdot du/dx \cdot dx = u \cdot du/2$. Using integration by parts with $u$ and $v$, we get:

$$\int u^2 \log(u) \cdot \frac{du}{2} = \frac{u^3}{3} \log(u) - \int \frac{u^3}{3} \cdot \frac{1}{u} \cdot \frac{du}{2}$$

$$= \frac{u^3}{3} \log(u) - \frac{1}{6} \int u^2 du = \frac{u^3}{3} \log(u) - \frac{u^3}{9} + C$$

Substituting back $u=2x+3$, we obtain the final result:

$$\int (2x+3)^2 \log(2x+3) dx = \frac{(2x+3)^3}{3} \log(2x+3) - \frac{(2x+3)^3}{9} + C$$