Question

Cum rezolv aceasta integrala?

Answer 1

[tex]\text{Folosesti integrarea prin parti:}\\ \displaystyle \int x\cdot \ln(x+1)dx=\displaystyle \int \left(\dfrac{x^2}{2}\right)'\cdot \ln(x+1) dx=\\= \dfrac{x^2}{2}\cdot \ln{(x+1)}-\displaystyle \int \dfrac{x^2}{2}\cdot (\ln{(x+1)})'dx=\\ = \dfrac{x^2\cdot \ln{(x+1)}}{2}-\displatyle\int \dfrac{x^2}{2(x+1)}dx \\ \text{Rezolvam a doua integrala:}\\ \int \dfrac{x^2}{2(x+1)}dx= \dfrac{1}{2}\int \dfrac{x^2}{x+1}dx\\ \text{Substituim:}\\ x+1=u\Rightarrow dx=du[/tex]
[tex]\displaystyle \dfrac{1}{2}\int \dfrac{(u-1)^2}{u}du=\dfrac{1}{2}\int \dfrac{u^2-2u+1}{u}du=\dfrac{1}{2}\left(\int udu-2\int du+\int \dfrac{du}{u}\right)\\ =\dfrac{1}{2}\cdot \left(\dfrac{u^2}{2}-2u+\ln u\right)=\dfrac{1}{2}\cdot \left(\dfrac{(x+1)^2}{2}-2(x+1)+\ln(x+1)\right)=\\ =\dfrac{(x+1)^2-4(x+1)+2\ln(x+1)}{4} \\ \text{Revenind:}\\ \boxed{ \int x\ln(x+1)dx= \dfrac{x^2\cdot \ln(x+1)}{2}-\dfrac{(x+1)^2-4(x+1)+2\ln(x+1)}{4}}\\ \text{Dupa ce inlocuiesti iti va da } \dfrac{1}{4}. [/tex]