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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 9 Maths Textbook Solution.

Answers (1)

Answer:

-x e^{-x}+C

Given:

\int(x-1) e^{-x} d x

Hint:

  Using integration by parts & \int e^{-x} d x

Explanation:

Let  I=\int(x-1) e^{-x} d x

          \begin{aligned} &=\int x e^{-x} d x-\int e^{-x} d x \\ &=x \frac{e^{-x}}{-1}+\int(1) e^{-x} d x-\int e^{-x} d x \end{aligned}                                    \left[\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right]

          =-x e^{-x}+\int e^{-x} d x-\int e^{-x} d x                                       \left[\because \int e^{x} d x=e^{x}+C\right]

           =-x e^{-x}+C

 

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