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Question Number 36430 by prof Abdo imad last updated on 02/Jun/18

$$find\int \frac{ln(x+x^2)}{x^2}dx$$

Commented by abdo mathsup last updated on 03/Jun/18

$$letintegratebyparts{u}^{{}^{\prime }}=\frac{1}{x^2}andv=ln(x+x^2)$$$$I=-\frac{1}{x}ln(x+x^2)+\int \frac{1}{x}\frac{2x+1}{x+x^2}dx$$$$=-\frac{ln(x+x^2)}{x}+\int \frac{2x+1}{x^2(x+1)}dxbut$$$$\int \frac{2x+1}{x^2(x+1)}dx=\int \frac{2(x+1)-1}{x^2(x+1)}dx$$$$=2\int \frac{dx}{x^2}-\int \frac{dx}{x^2(x+1)}=\frac{-2}{x}-\int \frac{dx}{x^2(x+1)}$$$$F\left(x\right)=\frac{1}{x^2(x+1)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x+1}$$$$b={lim}_{x\to 0}{x}^{2}F\left(x\right)=1$$$$c={lim}_{x\to -1}(x+1)F\left(x\right)=1\Rightarrow$$$$F\left(x\right)=\frac{1}{x}+\frac{b}{x^2}+\frac{1}{x+1}$$$$F\left(1\right)=\frac{1}{2}=1+b+\frac{1}{2}\Rightarrow b=-1\Rightarrow$$$$F\left(x\right)=\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x+1}\Rightarrow$$$$\int F\left(x\right)dx=ln\mid x\mid +\frac{1}{x}-ln\mid x+1\mid +c\Rightarrow$$$$I=-\frac{ln(x^2+x)}{x}-\frac{2}{x}-ln\mid x\mid -\frac{1}{x}+ln\mid x+1\mid +c\Rightarrow$$$$I=-\frac{ln(x^2+x)}{x}-\frac{3}{x}-ln\mid x\mid +ln\mid x+1\mid +c.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

$$ln(x+x^2)\times \frac{-1}{x}-\int \frac{1}{x+x^2}\times (1+2x)\times \frac{-1}{x}dx$$$$\frac{-ln(x+x^2)}{x}+\int \frac{1+2x}{x^2(1+x)}dx$$$$I_2=\int \frac{1+2x}{x^2(1+x)}dx$$$$=\int \frac{1+x+x}{x^2(1+x)}dx$$$$=\int \frac{dx}{x^2}+\int \frac{dx}{x(1+x)}$$$$=\int x^{-2}dx+\int \frac{1+x-x}{x(1+x)}dx$$$$=\frac{-1}{x}+\int \frac{dx}{x}-\int \frac{dx}{1+x}$$$$=\frac{-1}{x}+ln\left(\frac{x}{1+x}\right)$$$$=-\frac{ln(x+x^2)}{x}-\frac{1}{x}+ln\frac{x}{1+x}$$$$$$

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