Correct Answer is : xlog⁡(x+1)−x+log⁡(x+1)+cx (x+1) - x + (x+1) + cxlog(x+1)−x+log(x+1)+c

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Integral Calculus

Question : 36 of 73

Marks: +1, -0

Evaluate

\int \log(x+1) dx

Solution:

Concept:

Integrate using integration by parts:

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

Explanation:

Let

u = \log(x+1)

and

dv = dx

Then

du = \frac{1}{x+1} dx

and

v = x

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

Now

\frac{x}{x+1} = 1 - \frac{1}{x+1}

, so

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

Thus

\int \log(x+1) dx = x\log(x+1) - \left(x - \log(x+1)\right) + C = x\log(x+1) - x + \log(x+1) + C

Answer:

x\log(x+1) - x + \log(x+1) + c

, which corresponds to option A.

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