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How do you integrate #int x^4lnx# by integration by parts method?

Answer 1

Christopher Allers

#int x^n lnx dx = 1/(n+1)(x^(n+1)lnx-x^(n+1)/(n+1))+C#

#d/(dx)(x^n lnx) = n x^(n-1)lnx+x^(n-1)# or

#(n+1)int x^n lnx dx = x^(n+1)lnx-int x^n dx#

#int x^n lnx dx = 1/(n+1)(x^(n+1)lnx-x^(n+1)/(n+1))+C#

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Answer 2

Charles Arocha

To integrate ( \int x^4 \ln(x) ) using integration by parts, let ( u = \ln(x) ) and ( dv = x^4 dx ).

Then, ( du = \frac{1}{x} dx ) and ( v = \frac{x^5}{5} ).

Using the integration by parts formula:

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

we have:

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

Simplify the integral:

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

Integrate ( \int x^4 , dx ):

[ = \frac{x^5 \ln(x)}{5} - \frac{x^5}{25} + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

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.