How do you integrate (x^3)ln(x)dx(x3)ln(x)dx?

2 Answers
Ratnaker Mehta
Jul 27, 2016

x^4/16(4lnx-1)+Cx416(4lnx−1)+C,

OR,

ln{K(x^4/e)^(x^4/16)}ln⎧⎪⎨⎪⎩K(x4e)x416⎫⎪⎬⎪⎭

Explanation:

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

intuvdx=uintvdx-int[(du)/dxintvdx]dx∫uvdx=u∫vdx−∫[dudx∫vdx]dx

We take u=lnx, v=x^3rArr(du)/dx=1/x,&, intvdx=x^4/4u=lnx,v=x3⇒dudx=1x,&,∫vdx=x44

Hence, I=intx^3lnxdx=x^4/4*lnx-int(1/x*x^4/4)dxI=∫x3lnxdx=x44⋅lnx−∫(1x⋅x44)dx

=(x^4lnx)/4-1/4intx^3dx=(x^4lnx)/4-x^4/16=x4lnx4−14∫x3dx=x4lnx4−x416

:. 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.!

Cesareo R.
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

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