How do you find the antiderivative of #int 1/(x^2+10x+29) dx#?

Question

Ezra Adams · Accepted Answer

See the answer below:

William Abdalla · Answer

undefined To find the antiderivative of $\int \frac{1}{x^2+10x+29} \, dx$, you can complete the square in the denominator to rewrite the expression:

$$
x^2 + 10x + 29 = (x + 5)^2 + 4
$$

Now, substitute $u = x + 5$, then $du = dx$:

$$
\int \frac{1}{(x + 5)^2 + 4} \, dx = \int \frac{1}{u^2 + 4} \, du
$$

Now, you can use a trigonometric substitution. Let $u = 2 \tan(\theta)$, then $du = 2 \sec^2(\theta) \, d\theta$. Rewrite the integral in terms of $\theta$:

$$
\int \frac{1}{u^2 + 4} \, du = \frac{1}{2} \int \frac{1}{\tan^2(\theta) + 1} \cdot 2 \sec^2(\theta) \, d\theta
$$

This simplifies to:

$$
\frac{1}{2} \int \sec^2(\theta) \, d\theta = \frac{1}{2} \tan(\theta) + C
$$

Now, you need to back-substitute for $x$:

$$
\frac{1}{2} \tan(\theta) + C = \frac{1}{2} \tan\left(\arctan\left(\frac{x+5}{2}\right)\right) + C
$$

$$
= \frac{1}{2} \left(\frac{x+5}{2}\right) + C = \frac{1}{4} (x + 5) + C = \frac{x}{4} + \frac{5}{4} + C
$$

So, the antiderivative of $\frac{1}{x^2+10x+29}$ is $\frac{x}{4} + \frac{5}{4} + C$, where $C$ is the constant of integration.