What is ∫{dx}{x(1+In x)ⁿ} equal to (n ≠ 1) ?

Correct Answer is : −1(n−1)(1+Inx)n−1+c-{1}{(n-1)(1+In x)ⁿ-1}}+c−(n−1)(1+Inx)n−11+c

Correct Answer is : −1(n−1)(1+Inx)n−1+c-{1}{(n-1)(1+In x)ⁿ-1}}+c−(n−1)(1+Inx)n−11+c

Test Index

UPSC NDA 2 2019 Mathematics Solved Paper

Question : 82 of 120

Marks: +1, -0

What is

∫\frac{dx}{x(1+In x)^{n}}

equal to (n ≠ 1) ?

Solution:

Concept:

Use the substitution

t = 1 + \ln x

and the power rule

\int t^a dt = \frac{t^{a+1}}{a+1} + C

where

a \neq -1

Explanation:

Let

t = 1 + \ln x

Differentiate:

\frac{d}{dx}(1 + \ln x) = \frac{1}{x}

, so

\frac{dx}{x} = dt

The integral becomes

\int \frac{dx}{x(1+\ln x)^n} = \int \frac{dt}{t^n} = \int t^{-n} dt

Using the power rule:

\int t^{-n} dt = \frac{t^{-n+1}}{-n+1} + C = \frac{t^{1-n}}{1-n} + C

Rewrite:

\frac{t^{1-n}}{1-n} = -\frac{t^{1-n}}{n-1}

Substitute back

t = 1 + \ln x

-\frac{(1+\ln x)^{1-n}}{n-1} = -\frac{1}{(n-1)(1+\ln x)^{n-1}} + C

Answer:

Option D:

-\frac{1}{(n-1)(1+\ln x)^{n-1}} + C

Go to Question:

Prev Question

Next Question