Question

$\int e^x(x+\sqrt{1+x^2}+\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}})dx=$

• A. $e^x\cos h^{-1}x+c$
• B. $\frac{e^x}{\sqrt{1+x^2}}+c$
• C. $e^x$ $sin{h}^{-1}x+c$
• D. $e^x(x+\sqrt{x^2+1})+c$

Answer 1

The correct option is D $e^x(x+\sqrt{x^2+1})+c$

Let $I=\int e^x\left[\right(x+\sqrt{x+x^2})+\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}]dx$
$=\int e^x\left[\right(x+\sqrt{1+x^2})+(1+\frac{x}{\sqrt{1+x^2}}\left)\right]dx$
It is in the form of
$\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx$
$=e^{x}f\left(x\right)+c$
$\therefore I=e^x(x+\sqrt{1+x^2}+c)$
Hence, $\int e^x(x+\sqrt{1+x^2}+1+\frac{1}{\sqrt{1+x^2}})dx=e^x(x+\sqrt{1+x^2})+c$