# How do you integrate #int x^3lnx# by integration by parts method?

Integration by parts takes the form:

Thus:

Simplifying the integral:

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Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

So IBP gives:

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To integrate ( \int x^3 \ln(x) ) using integration by parts, you would apply the formula:

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

Let ( u = \ln(x) ) and ( dv = x^3 , dx ). Then, differentiate ( u ) to get ( du ) and integrate ( dv ) to get ( v ).

[ du = \frac{1}{x} , dx ] [ v = \frac{1}{4}x^4 ]

Now, apply the formula:

[ \int x^3 \ln(x) , dx = \frac{1}{4}x^4 \ln(x) - \int \frac{1}{4}x^4 \cdot \frac{1}{x} , dx ]

Simplify and solve the integral:

[ \int x^3 \ln(x) , dx = \frac{1}{4}x^4 \ln(x) - \frac{1}{4} \int x^3 , dx ]

[ = \frac{1}{4}x^4 \ln(x) - \frac{1}{16}x^4 + C ]

Therefore, ( \int x^3 \ln(x) , dx = \frac{1}{4}x^4 \ln(x) - \frac{1}{16}x^4 + C ), where ( C ) is the constant of integration.

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