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How do I find the integral #intx^5*ln(x)dx# ?

Answer 1

William Abdalla

By Parts Integration,

#int x^5lnx dx=x^6/36(6lnx-1)+C#

Let's examine a few specifics.

Let #u=lnx# and #dv=x^5dx#. #Rightarrow du={dx}/x# and #v=x^6/6#

Through Parts-Based Integration

#int udv=uv-int vdu#,

we have

#int (lnx)cdot x^5dx=(lnx)cdot x^6/6-int x^6/6cdot dx/x#

Simplifying a little bit

#=x^6/6lnx-int x^5/6dx#

through the Power Rule,

#=x^6/6lnx-x^6/36+C#

by factoring out #x^6/36#,

#=x^6/36(6lnx-1)+C#

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Answer 2

Ezekiel Alexander

To find the integral of ( \int x^5 \ln(x) , dx ), you can use integration by parts. Let ( u = \ln(x) ) and ( dv = x^5 , dx ). Then, ( du = \frac{1}{x} , dx ) and ( v = \frac{1}{6}x^6 ). Applying the integration by parts formula:

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

Substituting the values:

[ \int x^5 \ln(x) , dx = \frac{1}{6}x^6 \ln(x) - \int \frac{1}{6}x^5 , dx ]

Solving the remaining integral:

[ \int \frac{1}{6}x^5 , dx = \frac{1}{6} \times \frac{1}{6}x^6 + C = \frac{1}{36}x^6 + C ]

Thus, the integral of ( \int x^5 \ln(x) , dx ) is:

[ \frac{1}{6}x^6 \ln(x) - \frac{1}{36}x^6 + C ]

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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.