Find int(1)/(6x^(2)+36 x+78)dx. Choose 1 answer: (A) (1)/(2)arctan((x+3)/(2))+C (B) (1)/(2)arcsin((x+3)/(2))+C (C) (1)/(12)arcsin((x+3)/(2))+C (D) (1)/(12)arctan((x+3)/(2))+C

Complete the Square: First, complete the square for the quadratic in the denominator. 6x^2 + 36x + 78 = 6(x^2 + 6x + 13) Now, complete the square for x^2 + 6x + 13. x^2 + 6x + 13 = (x + 3)^2 + 4 So, 6x^2 + 36x + 78 = 6((x + 3)^2 + 4)Factor Out and Simplify: Next, factor out the 6 from the denominator to simplify the integral. int(1)/(6x^2 + 36x + 78)dx = int(1)/(6((x + 3)^2 + 4))dx = (1/6) * int(1)/((x + 3)^2 + 4)dxMake Substitution: Now, let's make a substitution to make the integral look like the standard arctan integral form. Let u = x + 3, then du = dx. The integral becomes (1/6) * int(1)/(u^2 + 4)du.Apply Arctan Integral Formula: The integral of 1/(u^2 + a^2) is (1/a) * arctan(u/a) + C. Here, a^2 = 4, so a = 2. The integral becomes (1/6) * (1/2) * arctan(u/2) + C.Substitute Back: Substitute back for u to get the integral in terms of x. (1/6) * (1/2) * arctan((x + 3)/2) + C = (1/12) * arctan((x + 3)/2) + C.Check Answer Choices: Check the answer choices to see which one matches our result. The correct answer is (D) (1/12)arctan((x+3)/(2))+C.

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Find
int(1)/(6x^(2)+36 x+78)dx.
Choose 1 answer:
(A)
(1)/(2)arctan((x+3)/(2))+C
(B)
(1)/(2)arcsin((x+3)/(2))+C
(C)
(1)/(12)arcsin((x+3)/(2))+C
(D)
(1)/(12)arctan((x+3)/(2))+C

Find $\int \frac{1}{6 x^{2}+36 x+78} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2} \arctan \left(\frac{x+3}{2}\right)+C$ $\newline$ (B) $\frac{1}{2} \arcsin \left(\frac{x+3}{2}\right)+C$ $\newline$ (C) $\frac{1}{12} \arcsin \left(\frac{x+3}{2}\right)+C$ $\newline$ (D) $\frac{1}{12} \arctan \left(\frac{x+3}{2}\right)+C$

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Find derivatives of using multiple formulae

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Q. Find

\int \frac{1}{6 x^{2}+36 x+78} d x

\newline

Choose

1

answer:

\newline

(A)

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

\newline

(B)

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

\newline

(C)

\frac{1}{12} \arcsin \left(\frac{x+3}{2}\right)+C

\newline

(D)

\frac{1}{12} \arctan \left(\frac{x+3}{2}\right)+C

Complete the Square: First, complete the square for the quadratic in the denominator. $\newline$ $6x^2 + 36x + 78 = 6(x^2 + 6x + 13)$ $\newline$ Now, complete the square for $x^2 + 6x + 13$ . $\newline$ $x^2 + 6x + 13 = (x + 3)^2 + 4$ $\newline$ So, $6x^2 + 36x + 78 = 6((x + 3)^2 + 4)$
Factor Out and Simplify: Next, factor out the $6$ from the denominator to simplify the integral. $\newline$ $\int \frac{1}{6x^2 + 36x + 78}\,dx = \int \frac{1}{6((x + 3)^2 + 4)}\,dx$ $\newline$ $= \frac{1}{6} \times \int \frac{1}{(x + 3)^2 + 4}\,dx$
Make Substitution: Now, let's make a substitution to make the integral look like the standard arctan integral form. $\newline$ Let $u = x + 3$ , then $du = dx$ . $\newline$ The integral becomes $(\frac{1}{6}) \int \frac{1}{u^2 + 4}du$ .
Apply Arctan Integral Formula: The integral of $\frac{1}{{u^2 + a^2}}$ is $\frac{1}{a} \cdot \text{arctan}\left(\frac{u}{a}\right) + C$ . Here, $a^2 = 4$ , so $a = 2$ . The integral becomes $\frac{1}{6} \cdot \frac{1}{2} \cdot \text{arctan}\left(\frac{u}{2}\right) + C$ .
Substitute Back: Substitute back for $u$ to get the integral in terms of $x$ . $\frac{1}{6} \times \frac{1}{2} \times \text{arctan}\left(\frac{x + 3}{2}\right) + C = \frac{1}{12} \times \text{arctan}\left(\frac{x + 3}{2}\right) + C$ .
Check Answer Choices: Check the answer choices to see which one matches our result. $\newline$ The correct answer is (D) $(\frac{1}{12})\arctan\left(\frac{x+3}{2}\right)+C$ .