Question Statement

Evaluate the integral:

$\int \frac{dy}{y^2+8y+20}$

---

Background and Explanation

This problem involves integrating a rational function with an irreducible quadratic denominator. The key technique is completing the square to transform the denominator into the form $u^2+a^2$, which allows us to use either a standard arctangent formula or trigonometric substitution.

---

Solution

First, we complete the square in the denominator:

\begin{align*} y^2 + 8y + 20 &= y^2 + 8y + 16 + 4 \\ &= (y+4)^2 + 4 \\ &= (y+4)^2 + 2^2 \end{align*}

So the integral becomes:

$I=\int \frac{dy}{(y+4{)}^2+2^2}$

Now we apply the trigonometric substitution. Let:

$y+4=2\tan \theta$

Taking differentials:

$dy=2{\sec }^2\theta d\theta$

Substituting into our integral:

\begin{align*} I &= \int \frac{2\sec^2\theta \, d\theta}{(2\tan\theta)^2 + (2)^2} \\ &= \int \frac{2\sec^2\theta \, d\theta}{4\tan^2\theta + 4} \\ &= \int \frac{2\sec^2\theta \, d\theta}{4(\tan^2\theta + 1)} \end{align*}

Using the Pythagorean identity ${\tan }^2\theta +1={\sec }^2\theta$:

\begin{align*} I &= \int \frac{2\sec^2\theta \, d\theta}{4\sec^2\theta} \\ &= \frac{1}{2} \int 1 \, d\theta \\ &= \frac{1}{2}\theta + C \end{align*}

To express the answer in terms of $y$, we back-substitute. Since $y+4=2\tan \theta$, we have:

$\tan \theta =\frac{y+4}{2}$

Therefore:

$\theta ={\tan }^{-1}\left(\frac{y+4}{2}\right)$

Substituting this back into our result:

$I=\frac{1}{2}{\tan }^{-1}\left(\frac{y+4}{2}\right)+C$

---

Key Formulas or Methods Used

• Completing the square: $y^2+by+c={\left(y+\frac{b}{2}\right)}^2+\left(c-\frac{b^2}{4}\right)$
• Trigonometric substitution: For integrals of the form $\frac{1}{u^2+a^2}$, use $u=a\tan \theta$
• Pythagorean identity: ${\tan }^2\theta +1={\sec }^2\theta$
• Standard integral form (alternative approach): $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$

---

Summary of Steps

1. Complete the square in the denominator: $y^2+8y+20=(y+4{)}^2+2^2$
2. Substitute $y+4=2\tan \theta$ and $dy=2{\sec }^2\theta d\theta$
3. Simplify the integrand using the identity ${\tan }^2\theta +1={\sec }^2\theta$ to reduce it to $\frac{1}{2}\int d\theta$
4. Integrate with respect to $\theta$ to get $\frac{1}{2}\theta +C$
5. Back-substitute $\theta ={\tan }^{-1}\left(\frac{y+4}{2}\right)$ to obtain the final answer in terms of $y$