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How do you integrate #int x^2lnx# by integration by parts method?

Answer 1

Charles Arocha

Use integration by parts. Let #u = lnx# and #dv = x^2dx# Then #du = 1/xdx# and #v = 1/3x^3#.

#intx^2lnx dx= 1/3x^3lnx - int(1/3x^3 * 1/x)dx#

#intx^2lnxdx = 1/3x^3lnx - int1/3x^2dx#

#intx^2lnx dx= 1/3x^3lnx - 1/9x^3 + C#

Hopefully this helps!

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Answer 2

Paisley Johnson

To integrate ∫x^2ln(x) using integration by parts, you can choose u = ln(x) and dv = x^2 dx. Then, differentiate u to find du and integrate dv to find v. Apply the integration by parts formula ∫u dv = uv - ∫v du, and substitute the values of u, v, du, and dv into the formula to evaluate the integral.

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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.