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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 5 Maths Textbook Solution.

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Answer: \left(\frac{x}{2}-\frac{1}{4}\right) e^{2 x}+c

Hint:Use integration by parts,

        \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x     

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

Solution: On using integration by parts

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

We Know that,\int e^{r x} d x=\frac{e^{r x}}{n} \& \frac{d}{d x} x=1

            \begin{aligned} &=\frac{x e^{2 x}}{2}-\int \frac{e^{2 x}}{2} d x \\ &=\frac{x e^{2 x}}{2}-\frac{e^{2 x}}{4}+c \\ &I=\left(\frac{x}{2}-\frac{1}{4}\right) e^{2 x}+c \end{aligned}

 

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