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Question Number 60318 by Sardor2211 last updated on 19/May/19

Commented by kaivan.ahmadi last updated on 20/May/19

$$wefindI=\int x^{2}arctgxdxthenmultiply4$$$$u=arctgx\Rightarrow du=\frac{dx}{1+x^2}$$$$dv=x^{2}dx\Rightarrow v=\frac{x^3}{3}$$$$I=uv-\int vdu=\frac{1}{3}{x}^{3}arctgx-\frac{1}{3}\int \frac{x^3}{1+x^2}dx=$$$$\frac{1}{3}{x}^{3}arctgx-\frac{1}{3}\int \frac{x^3+x-x}{1+x^2}dx=\frac{1}{3}{x}^{3}arctgx-$$$$\frac{1}{3}\int \frac{x^3+x}{1+x^2}dx+\frac{1}{3}\int \frac{x}{1+x^2}dx=\frac{1}{3}{x}^{3}arctgx-$$$$\frac{1}{3}\int xdx+\frac{1}{3}\int \frac{x}{1+x^2}dx=\frac{1}{3}{x}^{3}arctvx-\frac{1}{6}{x}^2+\frac{1}{6}ln(1+x^2)+C$$$$$$

Commented by maxmathsup by imad last updated on 20/May/19

$$letA=\int 4x^{2}arctan\left(x\right)dx\Rightarrow A=4\int x^{2}arctanxdxbyparts$$$$u^{{}^{\prime }}=x^{2}andv=arctanx\Rightarrow A=\frac{x^3}{3}arctanx-\int \frac{x^3}{3}\frac{dx}{1+x^2}$$$$=\frac{x^3}{3}arctan\left(x\right)-\frac{1}{3}\int \frac{x^3}{1+x^2}dx$$$$\int \frac{x^3}{1+x^2}dx=\int \frac{x(1+x^2)-x}{1+x^2}dx=\int xdx-\int \frac{x}{1+x^2}dx$$$$=\frac{x^2}{2}-\frac{1}{2}ln(1+x^2)+c\Rightarrow$$$$A=\frac{x^3}{3}arctan\left(x\right)-\frac{1}{6}{x}^2+\frac{1}{6}ln(1+x^2)+c.$$

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