# How do you integrate #int xln(1+x)# using integration by parts?

#I=1/2x^2ln(1+x)-1/2int(x-x/(1 +x))dx#

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To integrate ( \int x \ln(1+x) ) using integration by parts, you can apply the formula:

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

Let: [ u = \ln(1+x) ] [ dv = x , dx ]

Then: [ du = \frac{1}{1+x} , dx ] [ v = \frac{1}{2}x^2 ]

Using the formula: [ \int x \ln(1+x) , dx = \frac{1}{2}x^2 \ln(1+x) - \int \frac{1}{2}x^2 \frac{1}{1+x} , dx ]

Now integrate ( \frac{1}{2}x^2 \frac{1}{1+x} ) using a substitution or other methods.

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To integrate ( \int x \ln(1+x) ) using integration by parts, you can use the formula:

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

Choose ( u ) and ( dv ) such that it simplifies the integral. Let ( u = \ln(1+x) ) and ( dv = x , dx ). Then, differentiate ( u ) to find ( du ) and integrate ( dv ) to find ( v ).

[ u = \ln(1+x) ] [ dv = x , dx ]

Differentiating ( u ): [ du = \frac{1}{1+x} , dx ]

Integrating ( dv ): [ v = \frac{1}{2}x^2 ]

Now, apply the integration by parts formula:

[ \int x \ln(1+x) , dx = uv - \int v , du ] [ = \ln(1+x) \cdot \frac{1}{2}x^2 - \int \frac{1}{2}x^2 \cdot \frac{1}{1+x} , dx ]

Now, you have another integral to solve. You can either use further integration by parts or use other integration techniques like substitution or partial fractions to solve it.

This is the general procedure to integrate ( \int x \ln(1+x) ) using integration by parts.

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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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