# How do you integrate #int x^3ln(5x)# by integration by parts method?

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To integrate ( \int x^3 \ln(5x) ) using integration by parts, follow these steps:

- Choose ( u ) and ( dv ) for integration by parts. Typically, ( u ) is chosen as the function that becomes simpler when differentiated, and ( dv ) is chosen as the remaining part of the integrand.
- Compute ( du ) and ( v ) by differentiating ( u ) and integrating ( dv ), respectively.
- Apply the integration by parts formula:

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

- Substitute the computed values into the formula and simplify the integral.
- If necessary, repeat the process until you obtain a solvable integral.

For ( \int x^3 \ln(5x) ), let's choose ( u = \ln(5x) ) and ( dv = x^3 , dx ). Then,

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

Apply the integration by parts formula:

[ \int x^3 \ln(5x) , dx = \frac{x^4}{4} \ln(5x) - \int \frac{x^4}{4} \cdot \frac{1}{x} , dx ]

[ = \frac{x^4}{4} \ln(5x) - \frac{1}{4} \int x^3 , dx ]

Now, integrate ( \int x^3 , dx ) and simplify to obtain the final result.

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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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