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Find
int(1)/(x^(2)+10 x+41)dx.
Choose 1 answer:
(A)
(1)/(4)arctan((x+5)/(4))+C
B
(1)/(4)arcsin((x+5)/(4))+C
(C)
(1)/(20)arcsin((x+5)/(4))+C
(D)
(1)/(20)arctan((x+5)/(4))+C

Find $\int \frac{1}{x^{2}+10 x+41} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{4} \arctan \left(\frac{x+5}{4}\right)+C$ $\newline$ (B) $\frac{1}{4} \arcsin \left(\frac{x+5}{4}\right)+C$ $\newline$ (C) $\frac{1}{20} \arcsin \left(\frac{x+5}{4}\right)+C$ $\newline$ (D) $\frac{1}{20} \arctan \left(\frac{x+5}{4}\right)+C$

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Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{x^{2}+10 x+41} d x

.

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{4} \arctan \left(\frac{x+5}{4}\right)+C

\newline

(B)

\frac{1}{4} \arcsin \left(\frac{x+5}{4}\right)+C

\newline

(C)

\frac{1}{20} \arcsin \left(\frac{x+5}{4}\right)+C

\newline

(D)

\frac{1}{20} \arctan \left(\frac{x+5}{4}\right)+C

Complete the square: Complete the square for the denominator $x^2 + 10x + 41$ . $(x^2 + 10x + 25) + 16 = (x + 5)^2 + 4^2$
Rewrite the integral: Rewrite the integral using the completed square. $\int \frac{1}{(x + 5)^2 + 4^2} \, dx$
Recognize the integral: Recognize the integral as a form of the inverse tangent function. $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan(\frac{u}{a}) + C$ , where $u = x + 5$ and $a = 4$
Calculate the integral: Calculate the integral using the formula for arctan. $(\frac{1}{4}) \arctan(\frac{x + 5}{4}) + C$

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