Question

$\int e^{ta{n}^{-1}x}(1+x+x^2).d\left(co{t}^{-1}x\right)$ is equal to

• A. $-e^{ta{n}^{-1}x}+C$
• B. $e^{ta{n}^{-1}x}+C$
• C. $-x.e^{ta{n}^{-1}x}+C$
• D. $x.e^{ta{n}^{-1}x}+C$

Answer 1

The correct option is C

$-x.e^{ta{n}^{-1}x}+C$

$I=\int e^{ta{n}^{-1}x}(1+x+x^2).(-\frac{1}{1+x^2})dx$
$=-\int e^{ta{n}^{-1}x}(1+\frac{x}{1+x^2})dx$
$=-\int e^{ta{n}^{-1}x}dx-\int x.\frac{e^{ta{n}^{-1}x}}{1+x^2}dx$
$=-\int e^{ta{n}^{-1}x}dx-x.e^{ta{n}^{-1}x}+\int e^{ta{n}^{-1}x}dx+C$
$=-x.e^{ta{n}^{-1}x}+C$