Find int(1)/(x^(2)-6x+13)dx. Choose 1 answer: (A) (1)/(2)arcsin((x-3)/(2))+C (B) (1)/(4)arcsin((x-3)/(2))+C (C) (1)/(2)arctan((x-3)/(2))+C (D) (1)/(4)arctan((x-3)/(2))+C

Rewrite denominator: Rewrite the denominator as a complete square: x^2 - 6x + 13 = (x - 3)^2 + 4.Integral form: Now the integral looks like int(1)/((x-3)^2+4)dx.Recognize arctan derivative: Recognize the integral as a form of arctan derivative: int(1)/(a^2 + u^2)du = (1/a)arctan(u/a) + C.Substitute values: Here, a^2 = 4, so a = 2, and u = x - 3.Apply arctan formula: Substitute u and a into the arctan formula: int(1)/((x-3)^2+4)dx = (1/2)arctan((x-3)/2) + C.Match with answer: Match the result with the answer choices: (C) (1)/(2)arctan((x-3)/(2))+C is the correct one.

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Find
int(1)/(x^(2)-6x+13)dx.
Choose 1 answer:
(A)
(1)/(2)arcsin((x-3)/(2))+C
(B)
(1)/(4)arcsin((x-3)/(2))+C
(C)
(1)/(2)arctan((x-3)/(2))+C
(D)
(1)/(4)arctan((x-3)/(2))+C

Find $\int \frac{1}{x^{2}-6 x+13} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2} \arcsin \left(\frac{x-3}{2}\right)+C$ $\newline$ (B) $\frac{1}{4} \arcsin \left(\frac{x-3}{2}\right)+C$ $\newline$ (C) $\frac{1}{2} \arctan \left(\frac{x-3}{2}\right)+C$ $\newline$ (D) $\frac{1}{4} \arctan \left(\frac{x-3}{2}\right)+C$

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Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{x^{2}-6 x+13} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{2} \arcsin \left(\frac{x-3}{2}\right)+C

\newline

(B)

\frac{1}{4} \arcsin \left(\frac{x-3}{2}\right)+C

\newline

(C)

\frac{1}{2} \arctan \left(\frac{x-3}{2}\right)+C

\newline

(D)

\frac{1}{4} \arctan \left(\frac{x-3}{2}\right)+C

Rewrite denominator: Rewrite the denominator as a complete square: $x^2 - 6x + 13 = (x - 3)^2 + 4$ .
Integral form: Now the integral looks like $\int \frac{1}{(x-3)^2+4}\,dx$ .
Recognize arctan derivative: Recognize the integral as a form of arctan derivative: $\int \frac{1}{a^2 + u^2}du = \frac{1}{a}\arctan\left(\frac{u}{a}\right) + C$ .
Substitute values: Here, $a^2 = 4$ , so $a = 2$ , and $u = x - 3$ .
Apply arctan formula: Substitute $u$ and $a$ into the arctan formula: $\int \frac{1}{(x-3)^2+4}dx = \frac{1}{2}\arctan\left(\frac{x-3}{2}\right) + C$ .
Match with answer: Match the result with the answer choices: (C) $\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)+C$ is the correct one.