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Question Number 86956 by abdomathmax last updated on 01/Apr/20

$$find\int (1-\frac{1}{x^2})arctan\left(2x\right)dx$$

Commented by Ar Brandon last updated on 01/Apr/20

$$Letu=arctan\left(2x\right)\Rightarrow du=2\cdot \frac{1}{1+{\left(2x\right)}^2}dx$$$$dv=(1-\frac{1}{x^2})dx\Rightarrow v=(x+\frac{1}{x})$$$$$$$$I=(x+\frac{1}{x^2})arctan\left(2x\right)-2\int (x+\frac{1}{x})\cdot \left(\frac{1}{1+4x^2}\right)dx$$$$=(x+\frac{1}{x^2})arctan\left(2x\right)-2\int \frac{x^2+1}{x}\cdot \frac{1}{1+4x^2}dx$$$$=(x+\frac{1}{x^2})arctan\left(2x\right)-2\int \frac{x^2+1}{x(1+4x^2)}dx$$$$$$$$$$$$\frac{x^2+1}{x(1+4x^2)}=\frac{A}{x}+\frac{Bx+C}{4x^2+1}$$$$=\frac{A(4x^2+1)+(Bx+C)x}{x(4x^2+1)}=\frac{(4A+B)x^2+Cx+A}{x(4x^2+1)}$$$$4A+B=1$$$$C=0$$$$A=1\Rightarrow B=-3$$$$$$$$$$$$\int \frac{x^2}{x(1+4x^2)}dx=\int (\frac{1}{x}-\frac{3x}{1+4x^2})=\int \frac{1}{x}dx-\frac{3}{8}\int \frac{8x}{1+4x^2}dx$$$$=lnx-\frac{3}{8}ln(1+4x^2)+Constant$$$$$$$$\int (1-\frac{1}{x^2})arctan\left(2x\right)dx=(x+\frac{1}{x})arctan\left(2x\right)-2lnx+\frac{3}{4}ln(1+4x^2)+constant$$$$$$$$$$

Commented by mathmax by abdo last updated on 01/Apr/20

$$thankyousir$$

Commented by Ar Brandon last updated on 01/Apr/20

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