How do you integrate # (x^3)ln(x)dx#?

2 Answers
Jul 27, 2016

#x^4/16(4lnx-1)+C#,

OR,

#ln{K(x^4/e)^(x^4/16)}#

Explanation:

We use the Rule of Integration by Parts, which states,

#intuvdx=uintvdx-int[(du)/dxintvdx]dx#

We take #u=lnx, v=x^3rArr(du)/dx=1/x,&, intvdx=x^4/4#

Hence, #I=intx^3lnxdx=x^4/4*lnx-int(1/x*x^4/4)dx#

#=(x^4lnx)/4-1/4intx^3dx=(x^4lnx)/4-x^4/16#

#:. I=x^4/16(4lnx-1)+C#

#I# can further be shortened as under :-

#I=x^4/16(4lnx-1)=x^4/16(lnx^4-lne)=x^4/16ln(x^4/e)#

#:. I=ln(x^4/e)^(x^4/16)+lnK=ln{K(x^4/e)^(x^4/16)}#

Enjoy Maths.!

Jul 27, 2016

#1/4 x^4 log_e x-x^4/16 #

Explanation:

#d/(dx)(x^{n+1}log_e x)=(n+1)x^n log_e x+x^n#

so

#int x^n log_e x dx = 1/(n+1)x^{n+1}log_e x-int x^n/(n+1)dx+C#

Finally

#int x^n log_e x dx = 1/(n+1)x^{n+1}log_e x-x^{n+1}/(n+1)^2+C#

for #n = 3# we have the result

#1/4 x^4 log_e x-x^4/16 +C#