Question

Find the integral. $$ \int \frac{4}{1+9 x^{2}} d x $$

Short answer

The integral of \(\frac{4}{1+9 x^{2}}\) with respect to \(x\) is \(\frac{4}{3} x + C\). Where C is the constant of integration.

Step-by-step solution

Set up the substitution

Let's substitute \(x = \frac{1}{3} \tan(\theta)\). This implies that \(dx = \frac{1}{3}\sec^2(\theta) d\theta\).

Substitute into the integral

Substituting these into the integral, we get \(\int \frac{4}{1+9(\frac{1}{3}\tan(\theta))^2} \cdot \frac{1}{3}\sec^2(\theta) d\theta\) which simplifies to \(\int \frac{4}{3} \sec^2(\theta)d\theta\).

Evaluate the integral

The integral of \(\sec^2(\theta)\) is well known to be \(\tan(\theta)\), so integrating gives us \(\frac{4}{3} \tan(\theta) + C\).

Substitute back to original variable

Substitute back to the original variable \(x\). We have \(x = \frac{1}{3} \tan(\theta)\) which implies \(\theta = \arctan(3x)\). This gives the final answer as \(\frac{4}{3} \tan(\arctan(3x)) + C\), which simplifies to \(\frac{4}{3} x + C\).