Evaluate { x}{x²} \, dx

Correct Answer is : −1x(log⁡x+1)+c-1/x( x+1)+cx−1(logx+1)+c

Correct Answer is : −1x(log⁡x+1)+c-1/x( x+1)+cx−1(logx+1)+c

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Integral Calculus

Question : 33 of 73

Marks: +1, -0

Evaluate

\int \frac{\log x}{x^{2}} \, dx

Solution:

Concept:

Use integration by parts, treating

\log x

as the first function and

\frac{1}{x^2}

as the second function.

Explanation:

Let

I = \int \frac{\log x}{x^2} \, dx

Choose

u = \log x

and

dv = \frac{1}{x^2} dx

Then

du = \frac{1}{x} dx

and

v = \int x^{-2} dx = -\frac{1}{x}

Using integration by parts:

\int u \, dv = uv - \int v \, du

I = (\log x)\left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right)\left(\frac{1}{x}\right) dx

= -\frac{\log x}{x} + \int \frac{1}{x^2} dx

= -\frac{\log x}{x} + \left(-\frac{1}{x}\right) + c

= -\frac{1}{x}(\log x + 1) + c

Answer:

The integral evaluates to

-\frac{1}{x}(\log x + 1) + c

, which matches option B.

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