( x)^{-1} dx - ( x)^{-2} dx is equal to

Correct Answer is : x(ln⁡x)−1+cx ( x)^{-1} + cx(lnx)−1+c

Correct Answer is : x(ln⁡x)−1+cx ( x)^{-1} + cx(lnx)−1+c

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 88 of 120

Marks: +1, -0

\int (\ln x)^{-1} dx - \int (\ln x)^{-2} dx

is equal to

Solution:

Concept:

This problem simplifies a difference of two integrals using substitution and integration by parts to obtain a single closed form.

Explanation:

Let

I = \int \frac{1}{\ln x} \, dx - \int \frac{1}{(\ln x)^2} \, dx

Combine the integrals:

I = \int \left( \frac{1}{\ln x} - \frac{1}{(\ln x)^2} \right) dx

Use the substitution

t = \ln x

, so

x = e^t

and

dx = e^t \, dt

Then

I = \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt = \int \frac{e^t}{t} \, dt - \int \frac{e^t}{t^2} \, dt

Apply integration by parts to

\int \frac{e^t}{t} \, dt

: set

u = \frac{1}{t}

dv = e^t dt

, giving

du = -\frac{1}{t^2} dt

v = e^t

Then

\int \frac{e^t}{t} dt = \frac{e^t}{t} - \int e^t \left( -\frac{1}{t^2} \right) dt = \frac{e^t}{t} + \int \frac{e^t}{t^2} dt

Substitute this back into

I

I = \left( \frac{e^t}{t} + \int \frac{e^t}{t^2} dt \right) - \int \frac{e^t}{t^2} dt = \frac{e^t}{t} + C

Finally, replace

t = \ln x

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

Answer:

x (\ln x)^{-1} + c

(Option A).

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