# How do you integrate #(ln(x+1)/(x^2)) dx#?

by IBP this becomes:

so some partial fractions on this integral

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To integrate (\frac{\ln(x+1)}{x^2} , dx), you can use integration by parts. Let (u = \ln(x+1)) and (dv = \frac{1}{x^2} , dx). Then, (du = \frac{1}{x+1} , dx) and (v = -\frac{1}{x}).

Using the integration by parts formula: [\int u , dv = uv - \int v , du]

Substitute the values: [= -\ln(x+1) \cdot \frac{1}{x} - \int \left(-\frac{1}{x}\right) \cdot \frac{1}{x+1} , dx]

[= -\frac{\ln(x+1)}{x} + \int \frac{1}{x(x+1)} , dx]

For the integral (\int \frac{1}{x(x+1)} , dx), perform partial fraction decomposition: [\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}] [1 = A(x+1) + Bx]

Solve for (A) and (B): [A = 1] [B = -1]

So, (\int \frac{1}{x(x+1)} , dx = \ln|x| - \ln|x+1|)

Substitute this back into the original integration: [= -\frac{\ln(x+1)}{x} + \ln|x| - \ln|x+1| + C]

Thus, the integral (\int \frac{\ln(x+1)}{x^2} , dx) is: [= \ln|x| - \frac{\ln(x+1)}{x} - \ln|x+1| + C]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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