The integral of one over x^2+1

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Say we want to find

$\displaystyle I=\int{{\frac{{dx}}{{{{x}^{2}}+1}}}}$

If we happen to know that

$\displaystyle {{\left( {{{{\tan }}^{{-1}}}x} \right)}^{\prime }}=\frac{1}{{{{x}^{2}}+1}}$

then we have the answer. But suppose we don’t. There is a standard trigonometric substitution we could use, but we shall look at solving it here using a bit of complex analysis. For this particular problem, it is not a recommended way to do it. It is mainly just for fun, but as we will see with other problems, it can sometimes be a very useful method.

We start by doing the same as on the previous page. We will factorise the denominator.

$\displaystyle \int{{\frac{{dx}}{{{{x}^{2}}+1}}}}=\int{{\frac{{dx}}{{(x+i)(x-i)}}}}$

We can solve using partial fractions. We should find values A and B such that

$\displaystyle \frac{1}{{(x+i)(x-i)}}=\frac{A}{{x+i}}+\frac{B}{{x-i}}$

We multiply all of this by the denominator of the LHS to get

$\displaystyle 1=A(x-i)+B(x+i)$

Now, if we let x=i, we get 1=2iB, or B=1/2i. Now, let x= –i to get 1= –2iA or A= –1/2i. This gives us

$\displaystyle \begin{array}{l}\int{{\frac{{dx}}{{{{x}^{2}}+1}}}}=\frac{1}{{2i}}\int{{\left( {\frac{{-1}}{{x+i}}+\frac{1}{{x-i}}} \right)}}dx\\\quad \quad \quad =\frac{1}{{2i}}\left( {-\ln \left| {x+i} \right|+\ln \left| {x-i} \right|} \right)+C\\\quad \quad \quad =\frac{1}{{2i}}\ln \left| {\frac{{x-i}}{{x+i}}} \right|+C=I\end{array}$

The next step is to try to rewrite the above to something recognisable. Let us try to make x our subject. Perhaps we can see something we recognise then. We will not care about C in this.

First, we can multiply by 2i and then turn it into an exponential function.

$\displaystyle \left| {\frac{{x-i}}{{x+i}}} \right|={{e}^{{2iI}}}$

$\displaystyle \frac{{x-i}}{{x+i}}=\overbrace{{\pm {{e}^{{2iI}}}}}^{P}$

or,

$\displaystyle x-i=\left( {x+i} \right)\cdot P=xP+iP$

So we get

$\displaystyle x-xP=i+iP$

$\displaystyle x(1-P)=i(1+P)$

So,

$\displaystyle x=i\frac{{1+P}}{{1-P}}$

This will give us

$\displaystyle x=i\frac{{1\pm {{e}^{{2iI}}}}}{{1\mp {{e}^{{2iI}}}}}$

This looks suspiciously similar to a result found on this page.

$\displaystyle \tan \theta =\frac{1}{i}\cdot \frac{{{{e}^{{2i\theta }}}-1}}{{{{e}^{{2i\theta }}}+1}}$

In our expression for x we may first multiply both the denominator and the numerator by –1 to get:

$\displaystyle x=i\frac{{\pm {{e}^{{2iI}}}-1}}{{\mp {{e}^{{2iI}}}-1}}$

Then, since i= –1/i we can replace i by 1/i and change the sign of the denominator. This gives us

$\displaystyle x=\frac{1}{i}\cdot \frac{{\pm {{e}^{{2iI}}}-1}}{{\pm {{e}^{{2iI}}}+1}}$

If we select the positive sign on the ± we get the tan-function, so

$\displaystyle I={{\tan }^{{-1}}}x+C$

But how about if we choose the negative sign? That would give us

$\displaystyle x=\frac{1}{i}\cdot \frac{{-{{e}^{{2iI}}}-1}}{{-{{e}^{{2iI}}}+1}}=\frac{1}{i}\cdot \frac{{{{e}^{{2iI}}}+1}}{{{{e}^{{2iI}}}-1}}$

The reciprocal of that is

$\displaystyle \frac{1}{x}=i\cdot \frac{{{{e}^{{2iI}}}-1}}{{{{e}^{{2iI}}}+1}}=-\frac{1}{i}\cdot \frac{{{{e}^{{2iI}}}-1}}{{{{e}^{{2iI}}}+1}}=-\tan (I)$

$\displaystyle x=-{\cot}(I)$

So, another possible solution must be

$\displaystyle I={{\cot }^{{-1}}}(x)+C$

In the figure below, we can see those functions. I found it kind of neat that both solutions were true solutions.

Up a level : Integrals
Previous page : The intgral of one over x^2-1
Next page : The integral of 1/sqrt(1-x^2) dx

Last modified: May 22, 2025 @ 14:56