Use the substitution $$ x+2=2 an heta $$ to obtain $$ \int \frac{1}{x^{2}+4 x+8} d x $$

Question

Use the substitution $$ x+2=2 	an 	heta $$ to obtain $$ \int \frac{1}{x^{2}+4 x+8} d x $$

UpStudy AI · Accepted Answer

We start with the substitution
$$
x+2=2\tan\theta,
$$
and our goal is to evaluate
$$
\int \frac{1}{x^{2}+4x+8} \, dx.
$$

**Step 1. Express the quadratic in terms of $$ (x+2) $$.**

Notice that
$$
x^2 + 4x + 8 = (x+2)^2 + 4.
$$

Using the substitution $$ x+2=2\tan\theta $$, then
$$
(x+2)^2 = 4\tan^2\theta,
$$
so that
$$
x^2+4x+8 = 4\tan^2\theta + 4 = 4\left(\tan^2\theta+1\right) = 4\sec^2\theta.
$$

**Step 2. Compute $$ dx $$.**

Differentiate the substitution:
$$
\frac{d}{d\theta}(x+2)=\frac{d}{d\theta}(2\tan\theta) \quad \Longrightarrow \quad dx = 2\sec^2\theta \, d\theta.
$$

**Step 3. Substitute into the integral.**

Substitute $$ x^2+4x+8=4\sec^2\theta $$ and $$ dx=2\sec^2\theta \, d\theta $$:
$$
\int \frac{1}{4\sec^2\theta} \cdot \left(2\sec^2\theta\, d\theta\right) = \int \frac{2\sec^2\theta}{4\sec^2\theta} \, d\theta = \int \frac{1}{2}\, d\theta.
$$

**Step 4. Integrate with respect to $$ \theta $$.**

The integral simplifies to:
$$
\int \frac{1}{2}\, d\theta = \frac{1}{2}\theta + C.
$$

**Step 5. Express the result back in terms of $$ x $$.**

Recall the original substitution:
$$
x+2=2\tan\theta \quad \Longrightarrow \quad \tan\theta=\frac{x+2}{2}.
$$
Thus,
$$
\theta=\arctan\left(\frac{x+2}{2}\right).
$$

**Final Solution:**
$$
\int \frac{1}{x^{2}+4x+8} \, dx = \frac{1}{2}\arctan\left(\frac{x+2}{2}\right) + C.
$$
$$ \int \frac{1}{x^{2}+4x+8} \, dx = \frac{1}{2}\arctan\left(\frac{x+2}{2}\right) + C. $$

Use the substitution to obtain

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