\1/( x) - 1/(( x)^2)\ dx is equal to:(where C is an arbitrary constant) Option 1: 1/( x)+C Option 2: x^2/( x)+C Option 3: x/( x)+C Option 4: 1/(x x)+C

Answer: x/( x)+C. Solution: This is of the form e^f\g+g'\ style recognition; simplest is to differentiate the candidate. d/dx(x/( x))=( x - 1)/(( x)^2)=1/( x)-1/(( x)^2). This is exactly the integrand, so the integral is x/( x)+C.

Calculus

Integrals

Medium

common

$\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx$ is equal to:(where C is an arbitrary constant)

$\frac{1}{\log x}+C$

$\frac{x^2}{\log x}+C$

$\frac{x}{\log x}+C$

$\frac{1}{x\log x}+C$

✅ Correct Option: 3

This is of the form $\int e^{f}\{g+g'\}$ style recognition; simplest is to differentiate the candidate.

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

This is exactly the integrand, so the integral is $\frac{x}{\log x}+C$ .

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$\int \frac{dx}{(1+5\sin^2 x)}$ is equal to

∫{1log⁡x−1(log⁡x)2}dx\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx∫{logx1​−(logx)21​}dx is equal to:(where C is an arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx$ is equal to:(where C is an arbitrary constant)