# How do you evaluate the integral #intx^nlnxdx#?

For

For

Employ integration in segments. Let

Then:

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To evaluate the integral (\int x^n \ln x , dx), you can use integration by parts. The integration by parts formula states:

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

For this integral, let ( u = \ln x ) and ( dv = x^n , dx ). Then, differentiate ( u ) to find ( du ) and integrate ( dv ) to find ( v ). The steps are as follows:

- Let ( u = \ln x ) and ( dv = x^n , dx ).
- Differentiate ( u ) to find ( du ): ( du = \frac{1}{x} , dx ).
- Integrate ( dv ) to find ( v ): ( v = \frac{x^{n+1}}{n+1} ).
- Apply the integration by parts formula:

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

Now integrate ( \int x^n , dx ) separately:

[ \int x^n , dx = \frac{x^{n+1}}{n+1} + C ]

Substitute this back into the previous equation:

[ \int x^n \ln x , dx = \frac{x^{n+1} \ln x}{n+1} - \frac{1}{n+1} \left( \frac{x^{n+1}}{n+1} + C \right) ]

Simplify to get the final result:

[ \int x^n \ln x , dx = \frac{x^{n+1} \ln x}{n+1} - \frac{x^{n+1}}{(n+1)^2} - \frac{C}{n+1} ]

where ( C ) is the constant of integration.

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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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