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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 3 Maths Textbook Solution.

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Answer: \frac{x^{4}}{4} \log x-\frac{x^{4}}{16}+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: I=\int x^{3} \log x d x

Solution: On using Integration by parts

              I=\log x \int x^{3} d x-\int \frac{d}{d x} \log x \int x^{3} d x

 We have,\int x^{n} d x=\frac{x^{n+1}}{n+1} \& \frac{d}{d x} \log x=\frac{1}{x}

             \begin{aligned} &=\log x \times \frac{x^{4}}{4}-\int \frac{1}{x} \times \frac{x^{4}}{4} d x \\ &=\log x \times \frac{x^{4}}{4}-\frac{1}{4} \int x^{3} d x \\ &=\frac{x^{4}}{4} \log x-\frac{1}{4} \times \frac{x^{4}}{4} \\ &=\frac{x^{4}}{4} \log x-\frac{x^{4}}{16}+c \end{aligned}

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