How do you evaluate the integral of #int x^3 lnx dx#?
Use integration by parts, which states that:
Plugging these into the integration by parts formula, this yields:
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To evaluate the integral ∫x^3 ln(x) dx, you can use integration by parts. Let u = ln(x) and dv = x^3 dx.
Differentiate u to find du, and integrate dv to find v. Then, apply the integration by parts formula:
∫u dv = uv - ∫v du
Now, differentiate u = ln(x) to find du:
du = (1/x) dx
Integrate dv = x^3 dx to find v:
v = (1/4)x^4
Now, apply the integration by parts formula:
∫x^3 ln(x) dx = uv - ∫v du = (ln(x))(1/4)x^4 - ∫(1/4)x^4 (1/x) dx = (1/4)x^4 ln(x) - (1/4) ∫x^3 dx
Now, integrate ∫x^3 dx:
= (1/4)x^4 ln(x) - (1/4) * (1/4)x^4 + C = (1/4)x^4 ln(x) - (1/16)x^4 + C
So, the integral of ∫x^3 ln(x) dx is:
(1/4)x^4 ln(x) - (1/16)x^4 + C, 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.
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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