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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 6 Maths Textbook Solution.

Answers (1)

Answer: -e^{-x}\left(x^{2}+2 x+2\right)+c

Hint: Integration by parts says that:

        \begin{aligned} &\int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x \\ &u=x^{2}, \frac{d u}{d x}=2 x \\ &\frac{d v}{d x}=-e^{-x}, v=e^{-x} \\ &\int x^{2} e^{-x} d x=-x^{2} e^{-x}-\int-2 x e^{-2 x} d x \end{aligned}

Given: I=\int x^{2} e^{-x} d x

Solution: \int-2 x e^{-2 x} d x

             \begin{aligned} &u=2 x, \frac{d u}{d x}=2 \\ &\frac{d v}{d x}=-e^{-x}, v=e^{-x} \\ &=\int-2 x e^{-2 x} d x \\ &=2 x e^{-x}-\int 2 e^{-x} d x \\ &=2 x e^{-x}+2 e^{-x} \\ &=\int x^{2} e^{-x} d x \\ &=-x^{2} e^{-x}-\left(2 x e^{-x}+2 e^{-x}\right) \\ &=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+c \\ &=-e^{-x}\left(x^{2}+2 x+2\right)+c \end{aligned}

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