How do you integrate #(lnx)^2#?

Question

Cameron Alexander · Accepted Answer

First, note that #\int ln(x) dx# can be done by parts by letting #u=ln(x), dv=dx# so that #du=dx/x# and #v=x#, leading to #\int ln(x) dx=x ln(x)-\int dx=x ln(x)-x+C#. Now, for #\int (ln(x))^2 dx#, use integration-by-parts again with #u=ln(x)# and #dv=ln(x) dx# so that #du = dx/x# and #v=x ln(x)-x#.  Then #\int(\ln(x))^2 dx=x(ln(x))^2-x ln(x)-\int (ln(x)-1)dx=
x(ln(x))^2-x ln(x)+x-(x ln(x)-x)# #=x(ln(x))^2-2xln(x)+2x+C#.

Paisley Johnson · Answer

undefined To integrate $(\ln(x))^2$, you can use integration by parts. Let $u = \ln(x)$ and $dv = \ln(x) \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = x\ln(x) - \int x \cdot \frac{1}{x} \, dx$. Simplify $v$ to $v = x\ln(x) - \int dx$ and integrate to get $v = x\ln(x) - x$.

Now apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values:

$$ \int (\ln(x))^2 \, dx = x\ln(x) - x - \int x \cdot \frac{1}{x} \, dx $$

$$ \int (\ln(x))^2 \, dx = x\ln(x) - x - \int dx $$

$$ \int (\ln(x))^2 \, dx = x\ln(x) - x - x + C $$

$$ \int (\ln(x))^2 \, dx = x\ln(x) - 2x + C $$

So, the integral of $(\ln(x))^2$ is $x\ln(x) - 2x + C$, where $C$ is the constant of integration.