deksteenk9

2022-09-30

How to integrate $x\mathrm{ln}(x+1)$?

I am trying to compute $\int x\mathrm{ln}(x+1)\phantom{\rule{thinmathspace}{0ex}}dx$. I tried integrating by parts and ended up with:

$$\int x\mathrm{ln}(x+1)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2}{x}^{2}\mathrm{ln}(x+1)-\frac{1}{2}\int \frac{{x}^{2}}{x+1}\phantom{\rule{thinmathspace}{0ex}}dx$$

but I'm stuck here.

I am trying to compute $\int x\mathrm{ln}(x+1)\phantom{\rule{thinmathspace}{0ex}}dx$. I tried integrating by parts and ended up with:

$$\int x\mathrm{ln}(x+1)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2}{x}^{2}\mathrm{ln}(x+1)-\frac{1}{2}\int \frac{{x}^{2}}{x+1}\phantom{\rule{thinmathspace}{0ex}}dx$$

but I'm stuck here.

Demarion Thornton

Beginner2022-10-01Added 11 answers

Write ${x}^{2}={x}^{2}-1+1$ so that

$$\frac{{x}^{2}}{x+1}=\frac{{x}^{2}-1}{x+1}+\frac{1}{x+1}=(x-1)+\frac{1}{x+1}$$

$$\frac{{x}^{2}}{x+1}=\frac{{x}^{2}-1}{x+1}+\frac{1}{x+1}=(x-1)+\frac{1}{x+1}$$

Chasity Kane

Beginner2022-10-02Added 2 answers

$$\int {\displaystyle \frac{{x}^{2}}{x+1}}dx=\int {\displaystyle \frac{{x}^{2}-1}{x+1}}dx+\int {\displaystyle \frac{dx}{x+1}}=\int (x-1)dx+\int {\displaystyle \frac{dx}{x+1}}={\displaystyle \frac{{x}^{2}}{2}}-x+\mathrm{ln}(x+1)+c$$

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