How does one compute the integral of 1/(x^2+3)

Question

I’m trying to compute the integral of $$\frac{1}{x^2+3}$$ and one of the solutions is to use u substitution where $$u=\frac{x}{\sqrt{3}}$$ but I’m not sure how I would even come up with that substitution in the first place.

Answer 1

To solve this integral, you want to make use of the fact that

$$\int\frac{1}{x^2+1}\ {\rm d}x=\tan^{-1}(x)+C$$

This is probably the trickiest step, and needs some familiarity with common integrals. From there, we can use algebra to massage the integrand into the form $\frac{c}{u^2+1}$:

$$\frac{1}{x^2+3}=\frac{\frac{1}{3}}{\frac{x^2}{3}+1}=\frac{\frac{1}{3}}{\left(\frac{x}{\sqrt3}\right)^2+1}$$

Now from here, the $u$-substitution should be obvious. We can apply the u-substitution, $u=\frac{x}{\sqrt3},\,{\rm d}u=\frac{{\rm d}x}{\sqrt3}$ as follows:

$$\int{\frac{1}{x^2+3}\ {\rm d}x}=\int{\frac{\frac{1}{3}}{\left(\frac{x}{\sqrt3}\right)^2+1}\ {\rm d}x}\\ =\int{\frac{1}{3}\frac{1}{u^2+1}\sqrt3\ {\rm d}u}\\ =\frac{1}{\sqrt3}\int{\frac{1}{u^2+1}\ {\rm d}u}\\ =\frac{1}{\sqrt3}\tan^{-1}(u)+C\\ =\frac{1}{\sqrt3}\tan^{-1}(\frac{x}{\sqrt3})+C$$

Answer 2

Yes that substitution works. You end up getting $$ \int \frac{dx}{x^2+3} = \int \frac{1}{3u^2+3} \sqrt{3} \, du = \frac{1}{\sqrt{3}} \int \frac{du}{1+u^2} $$ Now recall that the antiderivative of $\frac{1}{1+u^2}$ is $\arctan u$.

Answer 3

There are $3$ principal "tracks" you can follow - there may be less obvious ones.

1. Trigonometric substitution. Try $x = \sqrt 3 \tan \theta$. This is the most accessible method with elementary knowledge. The substitution you asked about is strongly related to this method.

2. Hyperbolic Trigonometric substitution. Try $x = \sqrt 3 \sinh u$. A bit horrible since you have to know how to integrate $\mathrm{sech} u$.

3. Complex number partial fraction decomposition. Express the denominator as $(x + i\sqrt 3)(x - i\sqrt 3)$ and do a partial fraction decomposition. You'll have to eliminate the complex numbers with some post-integration algebra. Probably the fiddliest of the methods.

Answer 4

That substitution occurs to you because you remember that you know how to integrate $1/(1+x^2)$ (it's $\arctan$) and you figure out how to transform what you're given to what you know.

When you have worked out several similar integrals this will be a natural first step.

George Polya said

An idea which can be used only once is a trick. If one can use it more than once it becomes a method.

Answer 5

By substituting $$u=\frac{x}{\sqrt{3}}$$ we see that $$dx=\sqrt{3}du$$ So $$\int\frac{1}{x^2+3}dx=\int\frac{1}{(\sqrt{3}u)^2+3}\sqrt{3}du$$ By factoring out three from the denominator and moving constants outside of the integral, this can be simplified to $$ \frac{\sqrt{3}}{3}\int\frac{1}{u^2+1}du $$

The integrand is the derivative of $$\tan^{-1}{u}$$ so, after substituting back for $u$ and simplifying the constant in front, we get $$\frac{1}{\sqrt{3}}\tan^{-1}{\frac{x}{\sqrt{3}}}+C$$

Answer 6

Integration works better if you know a wide range of derivatives so that you know a wide range of antiderivatives.

• We know how to integrate polynomials because we know how to antidifferentiate polynomials because we know how to differentiate polynomials.
• We know how to integrate exponentials because ... we know how to differentiate exponentials.
• We know how to integrate trig functions because ... we know how to differentiate sine and cosine. For tangent, we differentiate using the quotient rule and antidifferentiate using substitution (the chain rule in reverse).
• We know how to integrate $1/x$ with respect to $x$ because we know what is the derivative of $\ln|x|$ with respect to $x$.
• We know how to integrate some rational functions because we know the derivatives of inverse (circular and hyperbolic) trig functions.

You want to integrate a rational function. So you should go look at the derivatives of inverse trig functions. We find: \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \arcsin(x) &= \frac{1}{\sqrt{1-x^2}} , x \in [-1,1] \\ \frac{\mathrm{d}}{\mathrm{d}x} \arccos(x) &= \frac{-1}{\sqrt{1-x^2}} , x \in [-1,1] \\ \frac{\mathrm{d}}{\mathrm{d}x} \arctan(x) &= \frac{1}{1+x^2} \\ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccot}(x) &= \frac{-1}{1+x^2} \\ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arcsec}(x) &= \frac{1}{|x|\sqrt{x^2-1}} , |x| > 1 \\ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccsc}(x) &= \frac{-1}{|x|\sqrt{x^2-1}} , |x| > 1 \\ \end{align*}

We know we can use substitution to absorb constant multiples of $x$ into the substitution variable (frequently, $u$). So we need a form that has the right pattern of constants, square roots, and powers of $x$, without getting too hung up on the values of the constants.

One bit of hard-won advice: make the constant term match by factoring first. This will cause some scaling of the other coefficients, but you can usually patch those up with your choice of $u$.

In your particular example, your pattern of signs, square roots, and constants is best matched by arctangent. We need to do a little algebra and a little substitution to get an exact match. \begin{align*} \frac{1}{x^2 + 3} &= \frac{1}{(\frac{1}{3}x^2 + 1)3} \\ &= \frac{1}{3} \cdot \frac{1}{\left(\frac{x}{\sqrt{3}}\right)^2 + 1} \text{.} \end{align*}

Now move the constant multiple, $1/3$ out front of your integral and set $u = x/\sqrt{3}$ so you get exactly $\displaystyle \frac{1}{1+u^2}$ as your integrand. (When you substitute for the $\mathrm{d}x$, you will pick up another constant multiple to move out front as well.)

If you would like to see a few more derivatives that can let you handle a few more integrands, see the derivatives of the inverse hyperbolic trig functions.