Question

$\int e^x\left[\frac{x^3+x+1}{(1+x^2{)}^{3/2}}\right]dx$ is equal to

• A. $\frac{x^2{e}^x}{(1+x^2{)}^{1/2}}+c$
• B. $\frac{e^x}{x(1+x^2{)}^{1/2}}+c$
• C. $\frac{e^x}{(1+x^2{)}^{1/2}}+c$
• D. $\frac{x{e}^x}{(1+x^2{)}^{1/2}}+c$

Answer 1

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

$I=\int e^x[\frac{x}{\sqrt{1+x^2}}+\frac{1}{(1+x^2{)}^{3/2}}]dx$

Let $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}}{(1+x^2)}$

$=\frac{1}{(1+x^2{)}^{3/2}}$

$\therefore I=e^{x}f\left(x\right)+c$

$=\frac{e^{x}x}{\sqrt{1+x^2}}+c$