# Find antiderivative of the function given by ln(x+1)/(x+1)?

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We have to find the anti derivative of [ln(x + 1)]/(x + 1)

The derivative of ln(x + 1) = 1/(x + 1)

Let ln(x + 1) = y

dy/dx = 1/(x + 1)

dy = dx/(x + 1)

Int [ [ln(x + 1)]/(x + 1) dx]

use ln(x+1) = y and substitute as shown earlier

=> Int [ y dy]

=> y^2 / 2 + C

Substitute y = ln (x + 1)

=> [ln( x + 1)]^2 / 2 + C

**The anti derivative of [ln(x + 1)]/(x + 1) = [ln( x + 1)]^2 / 2 + C**

To determine the antiderivative, we'll have to evaluate the indefinite integral of the function ln(x+1)/(x+1).

Int ln(x+1) dx/(x+1)

We'll substitute ln (x+1) by t.

ln(x+1) = y

We'll differentiate both sides:

dx/(x+1) = dy

We'll re-write the integral:

Int ydy = y^2/2 + C

Int ln(x+1) dx/(x+1) = [ln(x+1)]^2/2 + C

**The antiderivative of the function is Int ln(x+1) dx/(x+1) = [ln(x+1)]^2/2 + C.**