If f(x)=1/ x and g(x)={1}{( x)²}, then the value of [f(x)-g(x)] d x is...

Correct Answer is : xlog⁡x+cx/ x+clogxx+c

Correct Answer is : xlog⁡x+cx/ x+clogxx+c

Test Index

MHTCET 19 Apr 2026 Shift 1 Paper

Question : 130 of 150

Marks: +1, -0

f(x)=\frac{1}{\log x}

g(x)=\frac{1}{(\log x)^{2}}

\int[f(x)-g(x)] d x

is...

Solution:

Concept:

Identify the integrand as the derivative of

\frac{x}{\log x}

Explanation:

We are given

f(x)=\frac{1}{\log x}

and

g(x)=\frac{1}{(\log x)^2}

So,

f(x)-g(x)=\frac{1}{\log x}-\frac{1}{(\log x)^2}

Differentiate

\frac{x}{\log x}

using the quotient rule.

\frac{d}{dx}\left(\frac{x}{\log x}\right)=\frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2}

This simplifies to

\frac{\log x - 1}{(\log x)^2}

Rewriting,

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

Hence,

\int\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right]dx=\frac{x}{\log x}+c

Answer:

\frac{x}{\log x}+c

, which is Option C.

Go to Question:

Prev Question

Next Question