91Ó°ÊÓ

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 \(u(x) = 9 + x^2\).

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 \(d x\): $$ d x = \frac{d u}{2x} $$

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 \(u\): $$ \int \frac{1}{u^2} \cdot \frac{d u}{2\sqrt{u - 9}} $$

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: $$ \frac{1}{216} \arctan\left(\frac{x}{3}\right) + \frac{1}{432} \sin\left(2\arctan\left(\frac{x}{3}\right)\right) + \frac{1}{864} \left(1 + \cos\left(4\arctan\left(\frac{x}{3}\right)\right)\right) + C $$