Question

$\int e^{tan-1}x\left(\frac{1+x+x^2}{1+x^2}\right)dx=$

Answer 1

$\int e^{tan-1}x\left(\frac{1+x+x^2}{1+x^2}\right)dxLetI=\int e^{ta{n}^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)dx=\int e^{ta{n}^{-1}x}(\frac{1+x^2}{1+x^2}+\frac{x}{1+x^2})dx=\int e^{ta{n}^{-1}x}dx+\int \frac{x{e}^{ta{n}^{-1}x}}{1+x^2}dxI=I_1+I_{2}Now,I_2=\int \frac{x{e}^{ta{n}^{-1x}}}{1+x^2}dxPutta{n}^{-1}x=t\Rightarrow x=tant\Rightarrow \frac{1}{1+x^2}dx=dt\therefore I=\int \underset{I}{tant}.e^t\underset{II}{d}t$

$=tant.e^t-\int se{c}^{2}t.e^{t}dt+C=tant.e^t-\int (1+ta{n}^{2}t)e^{t}dt+C{I}_2=tant.e^t-\int (1+x^2)\frac{e^{tan-1x}}{1+x^2}dx+C{I}_2=tant.e^t-\int e^{tan-1x}dx+C=tant.e^t+C=x{e}^{ta{n}^{-1}x}+C$