How do you integrate #int x^3lnx# by integration by parts method?

2 Answers
mason m
Oct 25, 2016

#intx^3ln(x)dx=(x^4(4ln(x)-1))/16+C#

Explanation:

Integration by parts takes the form:

#intudv=uv-intvdu#

So, for the integral #intx^3ln(x)dx#, we see this is #intudv# and let:

#{(u=ln(x)" "=>" "du=1/xdx),(dv=x^3dx" "=>" "v=x^4/4):}#

Thus:

#intx^3ln(x)dx=uv-intvdu=(x^4ln(x))/4-intx^4/4 1/xdx#

Simplifying the integral:

#intx^3ln(x)dx=(x^4ln(x))/4-1/4intx^3dx#

#intx^3ln(x)dx=(x^4ln(x))/4-1/4x^4/4+C#

#intx^3ln(x)dx=(x^4ln(x))/4-x^4/16+C#

#intx^3ln(x)dx=(x^4(4ln(x)-1))/16+C#

Steve M
Oct 25, 2016

# int x^3lnxdx=1/4x^4lnx - x^4/16 + C #

Explanation:

The formula for integration by parts is:
# intu(dv)/dxdx = uv - intv(du)/dxdx #

Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

Let # {(u=lnx, => ,(du)/dx=1/x),((dv)/dx=x^3,=>,v =x^4/4 ):}#

So IBP gives:

# int (lnx)(x^3)dx=(lnx)(x^4/4) - int (x^4/4)(1/x)dx #
# :. int x^3lnxdx=1/4x^4lnx - 1/4int x^3dx #
# :. int x^3lnxdx=1/4x^4lnx - 1/4(x^4/4) + C #
# :. int x^3lnxdx=1/4x^4lnx - x^4/16 + C #

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