How do you integrate #int x^3lnx# by integration by parts method?
2 Answers
Explanation:
Integration by parts takes the form:
#intudv=uv-intvdu#
So, for the integral
#{(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#
Explanation:
The formula for integration by parts is:
Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).
Let
So IBP gives: