Get Answers to all your Questions

header-bg qa

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

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support