What is int ln(2x)∫ln(2x)?

2 Answers
Gió
Nov 4, 2015

I found: x(ln(2x)-1)x(ln(2x)−1)

Explanation:

You can try setting:
ln(2x)=tln(2x)=t so:
2x=e^t2x=et
x=1/2e^tx=12et
dx=1/2e^tdtdx=12etdt
The integral becomes:
intln(2x)dx=intte^t1/2dt=∫ln(2x)dx=∫tet12dt=
This can be solved using Integration by Parts:
=1/2[te^t-inte^t*1dt]==12[tet−∫et⋅1dt]=
=1/2[te^t-e^t]==12[tet−et]=
=1/2e^t(t-1)=12et(t−1)
Back to xx:
=(1/2)2x(ln(2x)-1)=x(ln(2x)-1)=(12)2x(ln(2x)−1)=x(ln(2x)−1)

Konstantinos Michailidis
Nov 4, 2015

You can go like this

int ln2xdx=int dx/dx*ln2xdx=xln2x-int x*dln(2x)/dxdx=xln2x-int x*2/(2x)dx=x*ln2x-x+c∫ln2xdx=∫dxdx⋅ln2xdx=xln2x−∫x⋅dln(2x)dxdx=xln2x−∫x⋅22xdx=x⋅ln2x−x+c

(*) We used integration by parts hence

int f'(x)*g(x)dx=f(x)*g(x)-int f(x)*g'(x)dx

where f(x)=x and g(x)=ln2x

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