The value of {xe}^{{x}} dx is equal to:

Correct Answer is : (2x−4x+4)ex+C(2 x-4 {x}+4) e^{{x}}+C(2x−4x+4)ex+C

Correct Answer is : (2x−4x+4)ex+C(2 x-4 {x}+4) e^{{x}}+C(2x−4x+4)ex+C

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UPSC NDA Math Model Paper 6

Question : 2 of 120

Marks: +1, -0

The value of

\int \sqrt{xe}^{\sqrt{x}} dx

Solution:

Let

I=\int \sqrt{xe}^{\sqrt{x}} dx

Substituting

\sqrt{x}=t

, we get:

\frac{1}{2\sqrt{x}} dx=dt

\Rightarrow dx=2t\,dt

\therefore I=\int t \times e^{t} \times 2t\,dt

Integrating by parts, taking

t^2

as the first function and

e^t

as the second function, we get:

\Rightarrow I=2\left[ t^{2} e^{t} dt - \int \left( \frac{d}{dt} t^{2} \int e^{t} dt \right) dt \right]+C

\Rightarrow I=2 t^{2} e^{t}-4 \int t e^{t} dt

+C

Integrating

\int t e^{t} dt

by parts, we get:

\Rightarrow I=2 t^{2} e^{t}-4\left[ t e^{t} dt - \int \left( \frac{d}{dt} t e^{t} dt \right) dt \right]

+C

\Rightarrow I=2 t^{2} e^{t}-4(t e^{t}-e^{t})+C

\Rightarrow I=(2 t^{2}-4 t+4) e^{t}+C

Back substituting

\sqrt{x}=t

, we get:

\Rightarrow I=(2 x-4 \sqrt{x}+4) e^{\sqrt{x}}+C

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