Integral of ln²(x)

In this post I will show you how to find the anti-derivative of the function f(x)=ln²(x).

$\int \ln^2(x)\,\mathrm{d}x$

We are going to use integration by parts:

$u=\ln^2(x)\;\;\;\;\;\mathrm{d}v=\mathrm{d}x$ $\mathrm{d}u=2\frac{\ln(x)}{x}\;\;\;\;\;v=x$

Now we can write u times v minus the integral of v times du:

$x\ln^2(x)-\int 2\frac{\ln(x)}{x}\,x\,\mathrm{d}x$

Let’s bring out the 2 (linearity); also, the two x‘s will cancel out.

$x\ln^2(x)-2\int \ln(x)\,\mathrm{d}x$

We saw here that the integral of ln(x) is equal to xln(x)-x, therefore:

$x\ln^2(x)-2\left ( x\ln(x)-x \right )=x\ln^2(x)-2x\ln(x)+2x$

If you want, you can factor out the x:

$x(\ln^2(x)-2\ln(x)+2)$

Finally,

$\int \ln^2(x)\,\mathrm{d}x=x(\ln^2(x)-2\ln(x)+2)+C$

That’s it!

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