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Integrating Inverse Tan

We're going to do some integration by parts!

The thing to realize is that you can integrate anything by parts. \(tan^{-1}(x)\) does indeed exist on its own. However, it is the same thing as \(tan^{-1}(x) \times 1\). So we'll integrate that expression by parts.

\(tan^{-1}(x) \times 1\)

The integral of 1 with respect to \(x\) is just \(x\).

\(tan^{-1}(x) \times \int (1)dx - \int (\frac{d}{dx}tan^{-1}(x) \times \int (1)dx) dx\)

Now, the derivative of \(tan^{-1}(x) is \frac{1}{1+x^2}\).

\(tan^{-1}(x) \times x - \int (\frac{1}{1+x^2} \times x)dx\)

\(xtan^{-1}(x) - \int (\frac{x}{1+x^2})dx \)

From here, there are two routes you can go down. You could make a substitution or use the \(\frac{f'(x)}{f(x)}\) property.

We'll do a substitution first.

Substitution

Let \(u=1+x^2\)

\(du=2x\ dx\)

\(dx=\frac{du}{2x}\)

Now that we have \(u\) and \(dx\), let's plug these values in.

\(xtan^{-1}(x) - \int (\frac{x}{u} \times \frac{du}{2x})\)

If we look at \(\frac{x}{u} \times \frac{du}{2x}\), we can simplify this to \(\frac{x}{2ux}du\). We will proceed to DESTROY and ANNIHILATE the \(x\) in this fraction.

\(xtan^{-1}(x) - \int (\frac{1}{2u} ) du\)

\(xtan^{-1}(x) - \frac{1}{2} \int (\frac{1}{u}) du\)

Using the \(\frac{f'(x)}{f(x)}\) property, we know that \(\frac{1}{u}du\) will integrate to \(ln(u)\). Hence:

\(xtan^{-1}(x) - \frac{1}{2} [ln(u)]\)

And since \(u=1+x^2\),

\(xtan^{-1}(x) - \frac{1}{2}ln(1+x^2)\)

That's it!

Using \(\frac{f'(x)}{f(x)}\)

This way is a fair bit shorter. Here's where we are:

\(xtan^{-1}(x) - \int (\frac{x}{1+x^2})dx \)

Focus on \(\frac{x}{1+x^2}\).

We know that the derivative of \(1+x^2\) is \(2x\). We can morph this value into the numerator if we multiply it by \(\frac{1}{2}\). Then, we have an example of \(\frac{f'(x)}{f(x)}\).

So, if we integrate \(\frac{x}{1+x^2}\), we end up with \(ln(1+x^2)\). If we plug that into the rest of our expression, we end up with \(xtan^{-1}(x) - \frac{1}{2}ln(1+x^2)\). That was a lot faster!

Think of it like \(\frac{1}{\frac{1}{2}} \times \frac{f'(x)}{2f(x)}\). This would simplify to \(2 \times \frac{f'(x)}{2f(x)}\) which becomes our coveted \(\frac{f'(x)}{f(x)}\).

This article was written on 30/01/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!