# How do you evaluate the integral #int x^-3lnx#?

The answer is

We do an integration by parts

Therefore,

By signing up, you agree to our Terms of Service and Privacy Policy

To evaluate the integral ( \int x^{-3}\ln(x) , dx ), you can use integration by parts. Let ( u = \ln(x) ) and ( dv = x^{-3} , dx ). Then, ( du = \frac{1}{x} , dx ) and ( v = -\frac{1}{2x^2} ). Applying the integration by parts formula, you get ( \int x^{-3}\ln(x) , dx = -\frac{\ln(x)}{2x^2} + \frac{1}{2} \int \frac{1}{x^3} , dx ). Integrate the remaining term ( \frac{1}{2x^3} ) to get ( -\frac{\ln(x)}{2x^2} - \frac{1}{4x^3} + C ), where ( C ) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7