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Question Number 170802 by Sotoberry last updated on 31/May/22

Answered by thfchristopher last updated on 31/May/22

$$\int \ln (x+x^2)dx$$$$=\int \ln x(x+1)dx$$$$=\int \ln xdx+\int \ln (x+1)dx$$$$=x\ln x-\int xd\left(\ln x\right)+x\ln (x+1)-\int xd\left[\ln (x+1)\right]$$$$=x\ln x-\int dx+x\ln (x+1)-\int \frac{x}{x+1}dx$$$$=x\ln x-x+x\ln (x+1)-\int (1-\frac{1}{x+1})dx$$$$=x\ln x-x+x\ln (x+1)-\int dx+\int \frac{1}{x+1}dx$$$$=x\ln x-2x+(x+1)\ln (x+1)+C$$$$=\ln x^x{(x+1)}^{x+1}-2x+C$$

Answered by Mathspace last updated on 31/May/22

$$bypartsI=xln(x+x^2)-\int x\times \frac{2x+1}{x+x^2}dx$$$$=xln(x+x^2)-\int \frac{2x+1}{x+1}dx$$$$=xln(x+x^2)-\int \frac{2(x+1)-1}{x+1}dx$$$$=xln(x+x^2)-2x+ln\mid x+1\mid +c$$

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