Integral of ln(2x)\ln\left(2 x\right)

The calculator will find the integral/antiderivative of ln(2x)\ln\left(2 x\right), with steps shown.

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Find ln(2x)dx\int \ln\left(2 x\right)\, dx.

Solution

Let u=2xu=2 x.

Then du=(2x)dx=2dxdu=\left(2 x\right)^{\prime }dx = 2 dx (steps can be seen »), and we have that dx=du2dx = \frac{du}{2}.

So,

ln(2x)dx=ln(u)2du{\color{red}{\int{\ln{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{2} d u}}}

Apply the constant multiple rule cf(u)du=cf(u)du\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du with c=12c=\frac{1}{2} and f(u)=ln(u)f{\left(u \right)} = \ln{\left(u \right)}:

ln(u)2du=(ln(u)du2){\color{red}{\int{\frac{\ln{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{2}\right)}}

For the integral ln(u)du\int{\ln{\left(u \right)} d u}, use integration by parts mdv=mvvdm\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}.

Let m=ln(u)\operatorname{m}=\ln{\left(u \right)} and dv=du\operatorname{dv}=du.

Then dm=(ln(u))du=duu\operatorname{dm}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u} (steps can be seen ») and v=1du=u\operatorname{v}=\int{1 d u}=u (steps can be seen »).

So,

ln(u)du2=(ln(u)uu1udu)2=(uln(u)1du)2\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{2}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{2}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{2}

Apply the constant rule cdu=cu\int c\, du = c u with c=1c=1:

uln(u)21du2=uln(u)2u2\frac{u \ln{\left(u \right)}}{2} - \frac{{\color{red}{\int{1 d u}}}}{2} = \frac{u \ln{\left(u \right)}}{2} - \frac{{\color{red}{u}}}{2}

Recall that u=2xu=2 x:

u2+uln(u)2=(2x)2+(2x)ln((2x))2- \frac{{\color{red}{u}}}{2} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{2} = - \frac{{\color{red}{\left(2 x\right)}}}{2} + \frac{{\color{red}{\left(2 x\right)}} \ln{\left({\color{red}{\left(2 x\right)}} \right)}}{2}

Therefore,

ln(2x)dx=xln(2x)x\int{\ln{\left(2 x \right)} d x} = x \ln{\left(2 x \right)} - x

Simplify:

ln(2x)dx=x(ln(x)1+ln(2))\int{\ln{\left(2 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(2 \right)}\right)

Add the constant of integration:

ln(2x)dx=x(ln(x)1+ln(2))+C\int{\ln{\left(2 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(2 \right)}\right)+C

Answer: ln(2x)dx=x(ln(x)1+ln(2))+C\int{\ln{\left(2 x \right)} d x}=x \left(\ln{\left(x \right)} - 1 + \ln{\left(2 \right)}\right)+C