Find int(1)/(x^(2)-14 x+58)dx. Choose 1 answer: (A) (1)/(21)arcsin((x-7)/(3))+C (B) (1)/(21)arctan((x-7)/(3))+C (C) (1)/(3)arctan((x-7)/(3))+C (D) (1)/(3)arcsin((x-7)/(3))+C

Complete the Square: First, let's complete the square for the denominator x^2 - 14x + 58. x^2 - 14x + 49 + 9 = (x - 7)^2 + 9 So, the integral becomes int(1)/((x-7)^2+9)dx.Use Substitution: Now, let's use a substitution to make it look like the arctan derivative. Let u = (x - 7)/3, then du = (1/3)dx, so dx = 3du.Substitute u and dx: Substitute u and dx into the integral. int(1)/((x-7)^2+9)dx = int(1)/((3u)^2+9) * 3du = int(1)/(9u^2+9) * 3du = int(1)/(9(u^2+1)) * 3du.Simplify the Integral: Simplify the integral. int(1)/(9(u^2+1)) * 3du = int(1)/(3(u^2+1))du.Recognize Integral: Recognize that the integral of 1/(u^2+1) is arctan(u). So, int(1)/(3(u^2+1))du = (1/3)arctan(u) + C.Substitute u back: Substitute u back in terms of x. (1/3)arctan(u) + C = (1/3)arctan((x-7)/3) + C.

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Find
int(1)/(x^(2)-14 x+58)dx.
Choose 1 answer:
(A)
(1)/(21)arcsin((x-7)/(3))+C
(B)
(1)/(21)arctan((x-7)/(3))+C
(C)
(1)/(3)arctan((x-7)/(3))+C
(D)
(1)/(3)arcsin((x-7)/(3))+C

Find $\int \frac{1}{x^{2}-14 x+58} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{21} \arcsin \left(\frac{x-7}{3}\right)+C$ $\newline$ (B) $\frac{1}{21} \arctan \left(\frac{x-7}{3}\right)+C$ $\newline$ (C) $\frac{1}{3} \arctan \left(\frac{x-7}{3}\right)+C$ $\newline$ (D) $\frac{1}{3} \arcsin \left(\frac{x-7}{3}\right)+C$

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Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{x^{2}-14 x+58} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{21} \arcsin \left(\frac{x-7}{3}\right)+C

\newline

(B)

\frac{1}{21} \arctan \left(\frac{x-7}{3}\right)+C

\newline

(C)

\frac{1}{3} \arctan \left(\frac{x-7}{3}\right)+C

\newline

(D)

\frac{1}{3} \arcsin \left(\frac{x-7}{3}\right)+C

Complete the Square: First, let's complete the square for the denominator $x^2 - 14x + 58$ . $\newline$ $x^2 - 14x + 49 + 9 = (x - 7)^2 + 9$ $\newline$ So, the integral becomes $\int \frac{1}{((x-7)^2+9)}dx$ .
Use Substitution: Now, let's use a substitution to make it look like the arctan derivative. $\newline$ Let $u = \frac{x - 7}{3}$ , then $du = \frac{1}{3}dx$ , so $dx = 3du$ .
Substitute $u$ and $dx$ : Substitute $u$ and $dx$ into the integral. $\int \frac{1}{(x-7)^2+9}dx = \int \frac{1}{(3u)^2+9} \cdot 3du = \int \frac{1}{9u^2+9} \cdot 3du = \int \frac{1}{9(u^2+1)} \cdot 3du.$
Simplify the Integral: Simplify the integral. $\int\frac{1}{9(u^2+1)} \cdot 3du = \int\frac{1}{3(u^2+1)}du$ .
Recognize Integral: Recognize that the integral of $\frac{1}{u^2+1}$ is $\arctan(u)$ . So, $\int \frac{1}{3(u^2+1)}du = \frac{1}{3}\arctan(u) + C$ .
Substitute $u$ back: Substitute $u$ back in terms of $x$ . $\frac{1}{3}$ arctan( $u$ ) + $C$ = $\frac{1}{3}$ arctan $\left(\frac{x-7}{3}\right)$ + $C$ .