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How do you find the integral of #7x^2ln(x) dx#?

Answer 1

Ezekiel Alexander

I would use Integretion by Parts:

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Answer 2

Charles Arocha

How: Use integration by parts. (Any integral of the form #ax^nlnx# can be done by parts.)

Details: #int7x^2ln(x)dx=7intx^2lnxdx#

Let #u=lnx# and #dv=x^2dx#

This makes #du=1/xdx# and #v=intx^2dx=(1/3)x^3# (We'll add the #+C# later.)

#intudv=uv-intvdu#

#7intx^2lnxdx=7[lnx(1/3)x^3-int(1/3)x^3*1/xdx]#

#=7[1/3x^3lnx-1/3intx^3/xdx]=7[1/3x^3lnx-1/3intx^2dx]#

#=7[1/3x^3lnx-1/3(1/3x^3)]+C#

#=7/3x^3lnx-7/9x^3+C#

General Notice that the solution can be made quite general. For example: with 5 instead of 3 the problem becomes #intx^5lnxdx# which can be solved by the same reasoning to get ; #1/6x^6lnx-1/36x^6+C#

Note: For even better understanding, check the answers by differentiating.

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Answer 3

Christopher Allers

To find the integral of 7x^2 ln(x) dx, you can use integration by parts. Let u = ln(x) and dv = 7x^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. Finally, substitute back the expressions for u, v, du, and dv to get the result.

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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.