Question

If $I=\int \frac{x^{2}-x+1}{x^{2}+1}.e^{cot^{-1}x}dx= A\left(x\right)e^{cot^{-1}x}+C,$ then $A(x)$ is equal to :

• A. $-x$
• B. $x$
• C. $\sqrt{1-x}$
• D. $\sqrt{1+x}$

Answer 1

Answer: (B)

Let $I=\int \frac{x^{2}-x+1}{x^{2}+1}.e^{cot^{-1}x}dx$
Put $x=cot\,t \Rightarrow -cosec^{2}\,dt=dx$
Now, $1 + cot^{2}\,t = cosec^{2}\,t$
$\therefore I=\int \frac{e^{t}\left(cot^{2}\,t-cot\,t+1\right)}{\left(1+cot^{2}\,t\right)}\left(-cosec^{2}\,t\right)dt$
$=- \int e^{t}\left(cosec^{2}\,t-cot\,t\right)dt$
$=\int e^{t}\left(cot\,t-coesc^{2}\,t\right)dt$
$=e^{t}\,cot\,t+C$
$=e^{cot^{-1}x}\left(x\right)+C\equiv A\left(x\right).e^{cot^{-1}x}+C$