Question

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

Answer: (C)

$x{e}^{ta{n}^{-1}x}+c$

$\int e^{ta{n}^{-1}x}(1+x+x^2)\cdot d\left(co{t}^{-1}x\right)=\int e^{ta{n}^{-1}x}\frac{(1+x+x^2)}{1+x^2}dx$
$=\int e^{ta{n}^{-1}x}dx+\int e^{ta{n}^{-1}x}\frac{x}{1+x^2}dx$
$=x{e}^{ta{n}^{-1}x}-\int \frac{x{e}^{ta{n}^{-1}x}}{1+x^2}dx+\int \frac{x{e}^{ta{n}^{-1}x}}{1+x^2}dx$ ..... Performing integration by parts for 1st integral
$=x{e}^{ta{n}^{-1}x}+c$