Value of (1/logx - 1/((logx)^2))dx is Option 1: x/logx + c, where c is an arbitrary constant Option 2: x logx + c, where c is an arbitrary constant Option 3: 1/logx + c, where c is an arbitrary constant Option 4: e^x/logx + c, where c is an arbitrary constant

Answer: x/logx + c, where c is an arbitrary constant. Solution: When an integral appears complicated, checking if any candidate function has the given expression as its derivative can be efficient. Consider x/( x) as a candidate for the antiderivative. Using the quotient rule where u = x and v = x: u' = 1 v' = 1/x d/dx(x/( x)) = ( x * (1) - x * (1/x))/(( x)^2) = ( x - 1)/(( x)^2) = ( x)/(( x)^2) - 1/(( x)^2) = 1/( x) - 1/(( x)^2) This matches the integrand exactly. Therefore: (1/( x) - 1/(( x)^2))dx = x/( x) + c

Calculus

Integrals

Medium

common

Value of $\int \left(\frac{1}{logx} - \frac{1}{(logx)^2}\right)dx$ is

$\frac{x}{logx} + c$ , where c is an arbitrary constant

$x\,logx + c$ , where c is an arbitrary constant

$\frac{1}{logx} + c$ , where c is an arbitrary constant

$\frac{e^x}{logx} + c$ , where c is an arbitrary constant

✅ Correct Option: 1

When an integral appears complicated, checking if any candidate function has the given expression as its derivative can be efficient.

Consider $\frac{x}{\log x}$ as a candidate for the antiderivative.

Using the quotient rule where $u = x$ and $v = \log x$ :

$u' = 1$

$v' = \frac{1}{x}$

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

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

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

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

This matches the integrand exactly.

Therefore: $\int \left(\frac{1}{\log x} - \frac{1}{(\log x)^2}\right)dx = \frac{x}{\log x} + c$

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Value of ∫(1logx−1(logx)2)dx\int \left(\frac{1}{logx} - \frac{1}{(logx)^2}\right)dx∫(logx1​−(logx)21​)dx is

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Value of $\int \left(\frac{1}{logx} - \frac{1}{(logx)^2}\right)dx$ is