Question

The value of integral $\int e^x(\frac{1}{\sqrt{1+x^2}}+\frac{1-2x^2}{\sqrt{(1+x^2{)}^5}})dx$ is equal to

• A. $e^x(\frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{(1+x^2{)}^3}})+c$
• B. $e^x(\frac{1}{\sqrt{1+x^2}}-\frac{x}{\sqrt{(1+x^2{)}^3}})+c$
• C. $e^x(\frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{(1+x^2{)}^5}})+c$
• D. none of these

Answer 1

The correct option is A $e^x(\frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{(1+x^2{)}^3}})+c$

$\int e^x(\frac{1}{\sqrt{1+x^2}}-\frac{x}{\sqrt{(1+x^2{)}^3}}+\frac{x}{\sqrt{(1+x^2{)}^3}}+\frac{1-2x^2}{\sqrt{(1+x^2{)}^5}})$
$=e^x\frac{1}{\sqrt{1+x^2}}+e^x\frac{x}{\sqrt{(1+x^2{)}^3}}=e^x(\frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{(1+x^2{)}^3}})+C$
Using $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx,$ we get
$e^{x}f\left(x\right)+c$

Ans: A