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Find
int(1)/(x^(2)+8x+52)dx.
Choose 1 answer:
(A)
(1)/(6)arcsin((x+4)/(6))+C
(B)
arctan((x+4)/(6))+C
(C)
(1)/(6)arctan((x+4)/(6))+C
(D)
arcsin((x+4)/(6))+C

Find $\int \frac{1}{x^{2}+8 x+52} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{6} \arcsin \left(\frac{x+4}{6}\right)+C$ $\newline$ (B) $\arctan \left(\frac{x+4}{6}\right)+C$ $\newline$ (C) $\frac{1}{6} \arctan \left(\frac{x+4}{6}\right)+C$ $\newline$ (D) $\arcsin \left(\frac{x+4}{6}\right)+C$

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Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{x^{2}+8 x+52} d x

.

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{6} \arcsin \left(\frac{x+4}{6}\right)+C

\newline

(B)

\arctan \left(\frac{x+4}{6}\right)+C

\newline

(C)

\frac{1}{6} \arctan \left(\frac{x+4}{6}\right)+C

\newline

(D)

\arcsin \left(\frac{x+4}{6}\right)+C

Rewrite with completed square: Rewrite the integral with the completed square. $\newline$ $\int \frac{1}{x^2 + 8x + 52} \, dx = \int \frac{1}{(x + 4)^2 + 6^2} \, dx$ .
Recognize inverse tangent form: Recognize the integral as a form of the inverse tangent function. $\newline$ The integral of $\frac{1}{a^2 + u^2} \, du$ is $\frac{1}{a} \cdot \arctan\left(\frac{u}{a}\right) + C$ . $\newline$ Here, $a = 6$ and $u = x + 4$ .
Calculate using inverse tangent formula: Calculate the integral using the formula for the inverse tangent. $\int \frac{1}{(x + 4)^2 + 6^2} dx = \frac{1}{6} \cdot \text{arctan}\left(\frac{x + 4}{6}\right) + C$ .

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