How do you evaluate the integral #int (lnx)^2dx#?

Question

Christopher Agresta · Accepted Answer

The answer is #=x(ln∣x∣)^2-2xln∣x∣+2x+C#

We carry out integration piece by piece.

#intuv'dx=uv-intu'vdx#

Here, #int(lnx)^2dx#

Let, #u=(lnx)^2#,#=>#, #u'=(2lnx)/x#

#v'=1#, #=>#, #v=x#

So,

#int(lnx)^2dx=x(lnx)^2-intx*(2lnx)/xdx#

#=x(lnx)^2-2intlnxdx#

We perform the partial integration once more.

#u=lnx#, #=>#, #u'=1/x#

#v'=1#, #=>#, #v=x#

so,

#intlnx=xlnx-intx*1/xdx#

#=xlnx-intdx=xlnx-x#

Combining everything at once

#int(lnx)^2dx=x(lnx)^2-2(xlnx-x)+C#

#=x(ln∣x∣)^2-2xln∣x∣+2x+C#

Christopher Agresta · Answer

undefined To evaluate the integral ∫(ln(x))^2 dx, you can use integration by parts. Let u = (ln(x))^2 and dv = dx. Then, du = 2(ln(x))(1/x) dx and v = x. Applying the integration by parts formula:

∫u dv = uv - ∫v du

We have:

∫(ln(x))^2 dx = x(ln(x))^2 - ∫x * 2(ln(x))(1/x) dx

Simplify the integral:

∫(ln(x))^2 dx = x(ln(x))^2 - 2∫ln(x) dx

Now integrate ∫ln(x) dx separately:

∫ln(x) dx = xln(x) - ∫x * (1/x) dx = xln(x) - ∫dx

This simplifies to:

∫ln(x) dx = xln(x) - x + C

Where C is the constant of integration. Substitute this back into the previous expression:

∫(ln(x))^2 dx = x(ln(x))^2 - 2(xln(x) - x) + C

Finally, simplify:

∫(ln(x))^2 dx = x(ln(x))^2 - 2xln(x) + 2x + C