# How do you integrate ln(x+1)?

Notice how you can write this as:

To integrate this, you can do Integration by Parts.

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To integrate (\ln(x+1)), you can use the technique of integration by parts. Let (u = \ln(x+1)) and (dv = dx). Then, differentiate (u) and integrate (dv) to find (du) and (v), respectively.

[u = \ln(x+1)] [du = \frac{1}{x+1}dx]

[dv = dx] [v = x]

Now, apply the integration by parts formula:

[ \int \ln(x+1) , dx = uv - \int v , du]

Substitute the values:

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

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

To solve the remaining integral, perform long division or use a technique like partial fractions. Once you've found the integral, you'll have the complete solution.

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

Using the integration by parts formula:

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

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

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

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

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

[ = x \ln(x+1) - x + \ln|x+1| + C ]

So, the integral of ( \ln(x+1) ) with respect to ( x ) is ( x \ln(x+1) - x + \ln|x+1| + C ), where ( C ) is the constant of integration.

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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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