How do you integrate #int x^nlnx# by integration by parts method?
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To integrate ( \int x^n \ln(x) ) using integration by parts, where ( n ) is a constant:

Choose ( u ) and ( dv ): Let ( u = \ln(x) ) and ( dv = x^n , dx ).

Compute ( du ) and ( v ): Differentiate ( u ) to get ( du = \frac{1}{x} , dx ), and integrate ( dv ) to get ( v = \frac{x^{n+1}}{n+1} ).

Apply the integration by parts formula: [ \int u , dv = uv  \int v , du ]

Substitute the values: [ \int x^n \ln(x) , dx = \ln(x) \cdot \frac{x^{n+1}}{n+1}  \int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} , dx ]

Simplify and integrate: [ \int x^n \ln(x) , dx = \frac{x^{n+1}}{n+1} \ln(x)  \frac{1}{n+1} \int x^n , dx ]

Finally, evaluate the integral ( \int x^n , dx ), which gives: [ \int x^n \ln(x) , dx = \frac{x^{n+1}}{n+1} \ln(x)  \frac{x^{n+1}}{(n+1)^2} + C ]
where ( C ) is the constant of integration.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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