Find int(1)/(2x^(2)-4x+20)dx. Choose 1 answer: (A) (1)/(3)arctan((x-1)/(3))+C (B) (1)/(6)arcsin((x-1)/(3))+C (C) (1)/(6)arctan((x-1)/(3))+C (D) (1)/(3)arcsin((x-1)/(3))+C

Complete the Square: First, complete the square for the quadratic in the denominator. 2x^2 - 4x + 20 = 2(x^2 - 2x + 10) Now, to complete the square, we need to add and subtract (b/2)^2 where b is the coefficient of x. So, (b/2)^2 = (2/2)^2 = 1. Add and subtract 1 inside the parenthesis and factor out the 2. 2(x^2 - 2x + 1 + 9) = 2((x - 1)^2 + 9)Rewrite the Integral: Now, rewrite the integral with the completed square. int(1)/(2x^2 - 4x + 20) dx = int(1)/(2((x - 1)^2 + 9)) dx Simplify the integral by taking out the constant. = (1/2) * int(1)/((x - 1)^2 + 9) dxRecognize Arctan Formula: Recognize the integral as a form of the inverse tangent function, arctan(u), where the integral of 1/(u^2 + a^2) du is (1/a) * arctan(u/a) + C. Here, u = x - 1 and a^2 = 9, so a = 3.Apply Arctan Formula: Apply the arctan formula. (1/2) * int(1)/((x - 1)^2 + 9) dx = (1/2) * (1/3) * arctan((x - 1)/3) + C Simplify the constants. = (1/6) * arctan((x - 1)/3) + C

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Find
int(1)/(2x^(2)-4x+20)dx.
Choose 1 answer:
(A)
(1)/(3)arctan((x-1)/(3))+C
(B)
(1)/(6)arcsin((x-1)/(3))+C
(C)
(1)/(6)arctan((x-1)/(3))+C
(D)
(1)/(3)arcsin((x-1)/(3))+C

Find $\int \frac{1}{2 x^{2}-4 x+20} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{3} \arctan \left(\frac{x-1}{3}\right)+C$ $\newline$ (B) $\frac{1}{6} \arcsin \left(\frac{x-1}{3}\right)+C$ $\newline$ (C) $\frac{1}{6} \arctan \left(\frac{x-1}{3}\right)+C$ $\newline$ (D) $\frac{1}{3} \arcsin \left(\frac{x-1}{3}\right)+C$

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Calculus

Find derivatives of using multiple formulae

Full solution

Q. Find

\int \frac{1}{2 x^{2}-4 x+20} d x

\newline

Choose

1

answer:

\newline

(A)

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

\newline

(B)

\frac{1}{6} \arcsin \left(\frac{x-1}{3}\right)+C

\newline

(C)

\frac{1}{6} \arctan \left(\frac{x-1}{3}\right)+C

\newline

(D)

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

Complete the Square: First, complete the square for the quadratic in the denominator. $\newline$ $2x^2 - 4x + 20 = 2(x^2 - 2x + 10)$ $\newline$ Now, to complete the square, we need to add and subtract $(b/2)^2$ where $b$ is the coefficient of $x$ . $\newline$ So, $(b/2)^2 = (2/2)^2 = 1$ . $\newline$ Add and subtract $1$ inside the parenthesis and factor out the $2$ . $\newline$ $2(x^2 - 2x + 1 + 9) = 2((x - 1)^2 + 9)$
Rewrite the Integral: Now, rewrite the integral with the completed square. $\newline$ $\int\frac{1}{2x^2 - 4x + 20} dx = \int\frac{1}{2((x - 1)^2 + 9)} dx$ $\newline$ Simplify the integral by taking out the constant. $\newline$ $= \frac{1}{2} \times \int\frac{1}{(x - 1)^2 + 9} dx$
Recognize Arctan Formula: Recognize the integral as a form of the inverse tangent function, $\arctan(u)$ , where the integral of $\frac{1}{u^2 + a^2} \, du$ is $\frac{1}{a} \cdot \arctan\left(\frac{u}{a}\right) + C$ . Here, $u = x - 1$ and $a^2 = 9$ , so $a = 3$ .
Apply Arctan Formula: Apply the arctan formula. $\newline$ $(\frac{1}{2}) \cdot \int \frac{1}{(x - 1)^2 + 9} dx = (\frac{1}{2}) \cdot (\frac{1}{3}) \cdot \arctan(\frac{x - 1}{3}) + C$ $\newline$ Simplify the constants. $\newline$ $= (\frac{1}{6}) \cdot \arctan(\frac{x - 1}{3}) + C$