Find int(1)/(3x^(2)+6x+78)dx. Choose 1 answer: (A) arctan((x+1)/(5))+C (B) arcsin((x+1)/(5))+C (c) (1)/(15)arctan((x+1)/(5))+C (D) (1)/(15)arcsin((x+1)/(5))+C

Complete the Square: First, let's complete the square for the quadratic in the denominator. 3x^2 + 6x + 78 = 3(x^2 + 2x + 26) Now, we need to add and subtract (2/2)^2 = 1 inside the parenthesis to complete the square. 3(x^2 + 2x + 1 - 1 + 26) 3((x + 1)^2 + 25)Rewrite with Completed Square: Now, we rewrite the integral with the completed square. int(1)/(3x^(2)+6x+78)dx = int(1)/(3((x+1)^2+25))dx Pull out the constant 3 from the denominator. = (1/3) * int(1)/((x+1)^2+25)dxApply Arctangent Integral Formula: Recognize that this is in the form of an arctangent integral. The integral of 1/(a^2 + u^2)du is (1/a) * arctan(u/a) + C, where a is a constant. Here, a^2 = 25, so a = 5.Integrate Using Arctangent Formula: Now, integrate using the arctangent formula. (1/3) * int(1)/((x+1)^2+25)dx = (1/3) * (1/5) * arctan((x+1)/5) + C Simplify the constants. = (1/15) * arctan((x+1)/5) + C

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Find
int(1)/(3x^(2)+6x+78)dx.
Choose 1 answer:
(A)
arctan((x+1)/(5))+C
(B)
arcsin((x+1)/(5))+C
(c)
(1)/(15)arctan((x+1)/(5))+C
(D)
(1)/(15)arcsin((x+1)/(5))+C

Find $\int \frac{1}{3 x^{2}+6 x+78} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\arctan \left(\frac{x+1}{5}\right)+C$ $\newline$ (B) $\arcsin \left(\frac{x+1}{5}\right)+C$ $\newline$ (c) $\frac{1}{15} \arctan \left(\frac{x+1}{5}\right)+C$ $\newline$ (D) $\frac{1}{15} \arcsin \left(\frac{x+1}{5}\right)+C$

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Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{3 x^{2}+6 x+78} d x

\newline

Choose

1

answer:

\newline

(A)

\arctan \left(\frac{x+1}{5}\right)+C

\newline

(B)

\arcsin \left(\frac{x+1}{5}\right)+C

\newline

(c)

\frac{1}{15} \arctan \left(\frac{x+1}{5}\right)+C

\newline

(D)

\frac{1}{15} \arcsin \left(\frac{x+1}{5}\right)+C

Complete the Square: First, let's complete the square for the quadratic in the denominator. $\newline$ $3x^2 + 6x + 78 = 3(x^2 + 2x + 26)$ $\newline$ Now, we need to add and subtract $(2/2)^2 = 1$ inside the parenthesis to complete the square. $\newline$ $3(x^2 + 2x + 1 - 1 + 26)$ $\newline$ $3((x + 1)^2 + 25)$
Rewrite with Completed Square: Now, we rewrite the integral with the completed square. $\newline$ $\int\frac{1}{3x^{2}+6x+78}\,dx = \int\frac{1}{3((x+1)^{2}+25)}\,dx$ $\newline$ Pull out the constant $3$ from the denominator. $\newline$ $= \frac{1}{3} \cdot \int\frac{1}{(x+1)^{2}+25}\,dx$
Apply Arctangent Integral Formula: Recognize that this is in the form of an arctangent integral. $\newline$ The integral of $\frac{1}{a^2 + u^2}du$ is $\frac{1}{a} \cdot \arctan(\frac{u}{a}) + C$ , where $a$ is a constant. $\newline$ Here, $a^2 = 25$ , so $a = 5$ .
Integrate Using Arctangent Formula: Now, integrate using the arctangent formula. $\newline$ $(\frac{1}{3}) \cdot \int \frac{1}{(x+1)^2+25}dx = (\frac{1}{3}) \cdot (\frac{1}{5}) \cdot \arctan(\frac{x+1}{5}) + C$ $\newline$ Simplify the constants. $\newline$ $= (\frac{1}{15}) \cdot \arctan(\frac{x+1}{5}) + C$