Question

$\int \frac{e^{ta{n}^{-1}x}}{(1+x^2)}[{\left(se{c}^{-1}\sqrt{1+x^2}\right)}^2+co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)]dx(x>0)$

• A. $e^{ta{n}^{-1}x}.ta{n}^{-1}x+C$
  
• B. $\frac{e^{ta{n}^{-1}x}.(ta{n}^{-1}x{)}^2}{2}+C$
  
• C. $e^{ta{n}^{-1}x}.{\left(se{c}^{-1}\right(\sqrt{1+x^2}\left)\right)}^2+C$
  
• D. $e^{ta{n}^{-1}x}.{\left(co{s}^{-1}\right(\sqrt{1+x^2}\left)\right)}^2+C$

Answer 1

The correct option is C $e^{ta{n}^{-1}x}.{\left(se{c}^{-1}\right(\sqrt{1+x^2}\left)\right)}^2+C$


note that $se{c}^{-1}\sqrt{1+x^2}=ta{n}^{-1}x;co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2ta{n}^{-1}x$ for x > 0
$I=\int \frac{e^{ta{n}^{-1x}}}{1+x^2}\left(\right(ta{n}^{-1}x{)}^2+2ta{n}^{-1}x)dx$
Put $ta{n}^{-1}x=t$
$=\int e^t(t^2+2t)dt=e^t.t^2=e^{ta{n}^{-1}x}\left(\right(ta{n}^{-1}x{)}^2)+C$