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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 15 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 15 Maths Textbook Solution.

Answers (1)

Answer: \frac{x^{-n+1}}{1-n} \log x-\frac{x^{-n+1}}{(1-n)^{2}}+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: \text { Let } I=\int \frac{\log x}{x^{n}} d x

Solution: I=\int \frac{\log x}{x^{n}} d x

               \begin{aligned} &=\int x^{-n} \log x d x \\ &=\log x \int x^{-n} d x-\int\left(\frac{d}{d x} \log x \int x^{-n} d x\right) d x \\ &=\log x\left(\frac{x^{-n+1}}{-n+1}\right)-\int \frac{1}{x} \cdot \frac{x^{-n+1}}{-n+1} d x \\ &=\frac{x^{-n+1} \log x}{1-n}-\frac{1}{1-n} \int \frac{x^{-n} x}{x} d x \end{aligned}

Again by integrating the second part

              =\frac{x^{-n+1} \log x}{1-n}-\frac{1}{1-n} \times \frac{x^{-n+1}}{-n+1}+c

So we get

                 =\frac{x^{-n+1}}{1-n} \log x-\frac{x^{-n+1}}{(1-n)^{2}}+c

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