Question

Evaluate the following integrals: $$ \int \frac{d x}{\left(9+x^{2}\right)^{2}} $$

Short answer

In order to evaluate the integral \(\int \frac{d x}{\left(9+x^{2}\right)^{2}}\), we first use substitution to rewrite the integral in terms of a new variable u: \(u = 9 + x^2\). This allows us to eliminate x terms from the integrand and make further simplifications. We then make another substitution, replacing u with t and using trigonometric identities to rewrite the integrand in terms of trigonometric functions. The integral is then evaluated and simplified in terms of θ, t, and finally x. The final result is: $$ \frac{1}{216} \arctan\left(\frac{x}{3}\right) + \frac{1}{432} \sin\left(2\arctan\left(\frac{x}{3}\right)\right) + \frac{1}{864} \left(1 + \cos\left(4\arctan\left(\frac{x}{3}\right)\right)\right) + C $$

Step-by-step solution

Identify the substitution

Observe the function inside the parentheses in the denominator: \(g(x) = 9 + x^2\). Our goal is to find a new function \(u(x)\), such that the integrand simplifies when we substitute \(u\) for this function. Let \(u(x) = 9 + x^2\).

Calculate the derivative of u(x)

Next, we calculate the derivative of \(u(x)\) with respect to \(x\). This will give us \(du\), which we can use to replace \(dx\) in the integral. $$ \frac{d u}{d x} = \frac{d}{d x} (9 + x^2) = 2x $$ Now, we solve for \(d x\): $$ d x = \frac{d u}{2x} $$

Write the integral in terms of u

Substitute \(u\) for \(9 + x^2\) and \(d x\) for \(\frac{d u}{2x}\) to rewrite the integral in terms of \(u\): $$ \int \frac{d x}{\left(9+x^{2}\right)^{2}} = \int \frac{1}{u^2} \cdot \frac{d u}{2x} $$ Notice that there's an \(x\) term in the numerator and the denominator. We can eliminate this term by solving the substitution equation \(u = 9 + x^2\) for \(x\): $$ x = \sqrt{u - 9} $$ Now, we can rewrite the integral fully in terms of \(u\): $$ \int \frac{1}{u^2} \cdot \frac{d u}{2\sqrt{u - 9}} $$

Simplify and solve the integral

We can cancel out the \(2\) in the denominator and the integral becomes: $$ \frac{1}{2} \int \frac{1}{u^2\sqrt{u - 9}} d u $$ Now, we can use the substitution \(t = \sqrt{u-9}\), which implies \(t^2 = u-9\) and therefore \(u=t^2+9\). The derivative of \(u\) with respect to \(t\) is: $$ \frac{d u}{d t} = 2t $$ Now, we solve for \(d u\): $$ d u = 2t d t $$ We substitute the new expression for \(u\) and \(d u\) and get: $$ \frac{1}{2} \int \frac{1}{(t^2+9)^2 \cdot t} (2t d t) $$ Now, we can cancel out the \(t\) term in the numerator and the denominator, and also the 2 in the numerator and the denominator: $$ \frac{1}{2} \int \frac{1}{(t^2+9)^2} d t $$ To solve this integral, we use the following trigonometric substitution: $$ t = 3 \tan(\theta) $$ Which implies: $$ d t = 3 \sec^2(\theta) d\theta $$ Now, substituting and applying the identity \(\tan^2(\theta) + 1 = \sec^2(\theta)\) in the integral: $$ \frac{1}{2} \int \frac{1}{(9 \tan^2(\theta) + 9)^2} \cdot 3 \sec^2(\theta) d\theta = \frac{3}{2} \int \frac{\sec^2(\theta)}{(9\sec^2(\theta))^2} d\theta $$ Simplifying and solving the integral, we get: $$ \frac{3}{2} \int \frac{1}{81\sec^4(\theta)} d\theta = \frac{1}{54} \int \cos^4(\theta) d\theta $$ To evaluate the integral, we use the following double-angle identity: $$ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} $$ So, \(\cos^4(\theta) = \left(\frac{1 + \cos(2\theta)}{2}\right)^2\). Replacing this in the integral: $$ \frac{1}{54} \int \left(\frac{1 + \cos(2\theta)}{2}\right)^2 d\theta $$ Expanding and combining terms: $$ \frac{1}{54} \int \left(\frac{1 + 2\cos(2\theta) + \cos^2(2\theta)}{4}\right) d\theta $$ Now, we can split the integral into separate parts: $$ \frac{1}{216} \int 1 d\theta + \frac{1}{108} \int \cos(2\theta) d\theta + \frac{1}{216} \int \cos^2(2\theta) d\theta $$ Evaluating these integrals, we get: $$ \frac{1}{216} \theta + \frac{1}{216} \left(\frac{\sin(2\theta)}{2}\right) + \frac{1}{432} \left(\frac{1 + \cos(4\theta)}{2}\right) + C $$ Now, we substitute back the expressions for \(\theta\) and \(t\): $$ \frac{1}{216} \arctan\left(\frac{t}{3}\right) + \frac{1}{432} \sin\left(2\arctan\left(\frac{t}{3}\right)\right) + \frac{1}{864} \left(1 + \cos\left(4\arctan\left(\frac{t}{3}\right)\right)\right) + C $$ Finally, we substitute back the expression for \(x\): $$ \frac{1}{216} \arctan\left(\frac{\sqrt{9+x^2 -9}}{3}\right) + \frac{1}{432} \sin\left(2\arctan\left(\frac{\sqrt{9+x^2 -9}}{3}\right)\right) + \frac{1}{864} \left(1 + \cos\left(4\arctan\left(\frac{\sqrt{9+x^2 -9}}{3}\right)\right)\right) + C $$ Simplifying the expression, we obtain the final result: $$ \frac{1}{216} \arctan\left(\frac{x}{3}\right) + \frac{1}{432} \sin\left(2\arctan\left(\frac{x}{3}\right)\right) + \frac{1}{864} \left(1 + \cos\left(4\arctan\left(\frac{x}{3}\right)\right)\right) + C $$