How do I solve an indefinite integral where dx is involved?

Question

I do not know how to evaluate the integral below, where $dx$ is in the numerator of the integrand:

$$I=\int \frac{dx}{9+x^2}$$

I would appreciate if somebody could walk me through the evaluation.

Answer 1

For your purposes, as I mentioned in the comments you can think of $dx$ as showing you that you are integrating with respect to $x$. Now for your integral we must perform a substitution.

We will use the substitution $x = 3\tan u$ which also yields that $dx/du = 3 \sec^2 u$

That means we can rewrite our integral like this

$$\int \frac{1}{9 + x^2} \frac{du}{dx}\frac{dx}{du}dx$$ $$\int \frac{1}{9 + (3\tan u)^2} 3\sec^2u\frac{du}{dx}dx$$ $$\int \frac{3\sec^2 u}{9(1 + \tan^2 u)} du$$ Now look how we are integrating with respect to $u$. Using the trig identity $1 + \tan^2u = \sec^2u$ $$\int \frac{1}{3}du = \frac{u}{3} + C$$

Now substituting $x$ back in $$\int \frac{1}{9 + x^2} dx = \frac{\arctan(\frac{x}{3})}{3} + C$$

Answer 2

HINT:

\begin{equation} \frac{d}{dx}(\mathrm{arctan}(ax)) = \frac{a}{1+(ax)^2} \end{equation}

$a\in\mathbb{R}$ is a fixed number.

Try to adapt your integral to use the hint.

Answer 3

I think your confusion is due to the $dx$ being within the fraction. Your integral is equivalent to: $$\int \frac{dx}{9+x^2}=\int \frac{1}{9+x^2}~dx$$

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To solve it, substitute: $$x=3\tan{\theta} \implies dx=3\sec^2{\theta}~d{\theta}$$ This gives: $$\int \frac{1}{9+x^2}~dx=\int \frac{1}{9+9\tan^2{\theta}}\cdot 3\sec^2{\theta}~d{\theta}$$ Now, use the identity: $$1+\tan^2{\theta}\equiv \sec^2{\theta}$$ Can you continue?