Question

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x})dx$ equals

• A. $2e^{\sqrt{x}}[x-\sqrt{x}+1]+C$
• B. $e^{\sqrt{x}}[x-2\sqrt{x}+1]$
• C. $e^{\sqrt{x}}[x+\sqrt{x}]+C$
• D. $e^{\sqrt{x}}[x+\sqrt{x}+1]+C$

Answer 1

The correct option is A $2e^{\sqrt{x}}[x-\sqrt{x}+1]+C$

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x})dx$;; Put $x=t^2;dx=2tdt$
$=\int e^t(t^2+t)dt=e^t(A{t}^2+Bt+C)$
Diffrentiate both the sides
$e^t(t^2+t)=e^t(2At+B)+(A{t}^2+Bt+C)e^t$
On comparing coefficient we get
A = 1 ; B = – 1 ; C = 1