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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!

cricket900
  • 185

2 Answers2

1

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:

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

Scientia
  • 29
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$I=\int u^2\ ln(u)\ du=\frac1{3}\int ln(u)\ du^3=\frac 1{3}[u^3\ ln(u)-\int u^3\ d(ln(u))]$

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

user295357
  • 505
  • Can you explain step by step what you did? – cricket900 Mar 18 '23 at 07:53
  • @ cricket900: First of all, if you do derivative of my result, you can easily verify that it is correct. The second and third equal signs in the first line are because of the method of integration by parts. The first equal sign in the second line is because $d(ln(u))=1/u\ du$ and $\int u^2\ du=\frac{1}{3}u^3$. – user295357 Mar 18 '23 at 15:42
  • For what reason someone gave my answer a -1 ? Why my answer is not useful ? Please give your explanation. If you don't understand my answer, you should ask me, just like cricket900 did above. If you think my answer is wrong, please tell me what is wrong. – user295357 Mar 18 '23 at 15:53
  • @cricket900: Can you see that Scientia's answer posted 35 minutes ago is same as my answer, simply with $u = 2x+3$ substituted into my result ? – user295357 Mar 18 '23 at 16:00