Find int(1)/(x^(2)-8x+65)dx. Choose 1 answer: (A) (1)/(7)arctan((x-4)/(7))+C (B) (1)/(7)arcsin((x-4)/(7))+C (C) arctan((x-4)/(7))+C (D) arcsin((x-4)/(7))+C

Complete the square: Complete the square for the denominator x^2 - 8x + 65 to make it look like (x - h)^2 + k^2, which is suitable for a trigonometric substitution. (x^2 - 8x + 16) + 49 = (x - 4)^2 + 7^2Recognize standard form: Recognize that the integral is now in the form of 1/((x - h)^2 + k^2), which is a standard form for arctan substitution.Use substitution u: Use the substitution u = (x - 4)/7, then du = (1/7)dx.Rewrite integral in terms: Rewrite the integral in terms of u: int(1)/(u^2 + 1) * 7du.Integrate using arctan formula: Integrate using the arctan formula: int(1)/(u^2 + 1)du = arctan(u) + C.Substitute back for x: Substitute back for x to get the final answer: (1/7)arctan((x - 4)/7) + C.

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Find
int(1)/(x^(2)-8x+65)dx.
Choose 1 answer:
(A)
(1)/(7)arctan((x-4)/(7))+C
(B)
(1)/(7)arcsin((x-4)/(7))+C
(C)
arctan((x-4)/(7))+C
(D)
arcsin((x-4)/(7))+C

Find $\int \frac{1}{x^{2}-8 x+65} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{7} \arctan \left(\frac{x-4}{7}\right)+C$ $\newline$ (B) $\frac{1}{7} \arcsin \left(\frac{x-4}{7}\right)+C$ $\newline$ (C) $\arctan \left(\frac{x-4}{7}\right)+C$ $\newline$ (D) $\arcsin \left(\frac{x-4}{7}\right)+C$

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Calculus

Find derivatives of using multiple formulae

Full solution

Q. Find

\int \frac{1}{x^{2}-8 x+65} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{7} \arctan \left(\frac{x-4}{7}\right)+C

\newline

(B)

\frac{1}{7} \arcsin \left(\frac{x-4}{7}\right)+C

\newline

(C)

\arctan \left(\frac{x-4}{7}\right)+C

\newline

(D)

\arcsin \left(\frac{x-4}{7}\right)+C

Complete the square: Complete the square for the denominator $x^2 - 8x + 65$ to make it look like $(x - h)^2 + k^2$ , which is suitable for a trigonometric substitution. $\newline$ $(x^2 - 8x + 16) + 49 = (x - 4)^2 + 7^2$
Recognize standard form: Recognize that the integral is now in the form of $\frac{1}{((x - h)^2 + k^2)}$ , which is a standard form for $\arctan$ substitution.
Use substitution $u$ : Use the substitution $u = \frac{x - 4}{7}$ , then $du = \frac{1}{7}dx$ .
Rewrite integral in terms: Rewrite the integral in terms of $u$ : $\int \frac{1}{u^2 + 1} \cdot 7\,du$ .
Integrate using arctan formula: Integrate using the arctan formula: $\int \frac{1}{u^2 + 1}\,du = \arctan(u) + C$ .
Substitute back for x: Substitute back for x to get the final answer: $(\frac{1}{7})\arctan(\frac{(x - 4)}{7}) + C$ .