Find int(1)/(sqrt(-x^(2)-14 x-48))dx. Choose 1 answer: (A) (1)/(2)arctan((x+7)/(2))+C (B) arcsin(x+7)+C (C) arctan(x+7)+C (D) (1)/(2)arcsin((x+7)/(2))+C

Complete the square: First, let's complete the square for the expression under the square root to make it look like a standard integral form. (-x^2 - 14x - 48) = -(x^2 + 14x + 49) + 1 = -(x + 7)^2 + 1Substitute u = x + 7: Now, let's substitute u = x + 7, then du = dx. So, the integral becomes int(1)/(sqrt(1 - u^2))du.Standard integral form: This is a standard integral form that resembles the inverse sine function, arcsin(u). So, the integral becomes arcsin(u) + C.Substitute back u: Substitute back u = x + 7 to get the final answer. arcsin(x + 7) + C.Match with options: Now, let's match our answer with the given options. The correct answer is (B) arcsin(x+7)+C.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find
int(1)/(sqrt(-x^(2)-14 x-48))dx.
Choose 1 answer:
(A)
(1)/(2)arctan((x+7)/(2))+C
(B)
arcsin(x+7)+C
(C)
arctan(x+7)+C
(D)
(1)/(2)arcsin((x+7)/(2))+C

Find $\int \frac{1}{\sqrt{-x^{2}-14 x-48}} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2} \arctan \left(\frac{x+7}{2}\right)+C$ $\newline$ (B) $\arcsin (x+7)+C$ $\newline$ (C) $\arctan (x+7)+C$ $\newline$ (D) $\frac{1}{2} \arcsin \left(\frac{x+7}{2}\right)+C$

View step-by-step help

Home

Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{\sqrt{-x^{2}-14 x-48}} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{2} \arctan \left(\frac{x+7}{2}\right)+C

\newline

(B)

\arcsin (x+7)+C

\newline

(C)

\arctan (x+7)+C

\newline

(D)

\frac{1}{2} \arcsin \left(\frac{x+7}{2}\right)+C

Complete the square: First, let's complete the square for the expression under the square root to make it look like a standard integral form. $\newline$ $(-x^2 - 14x - 48) = -(x^2 + 14x + 49) + 1 = -(x + 7)^2 + 1$
Substitute $u = x + 7$ : Now, let's substitute $u = x + 7$ , then $du = dx$ . So, the integral becomes $\int \frac{1}{\sqrt{1 - u^2}}\,du$ .
Standard integral form: This is a standard integral form that resembles the inverse sine function, $\arcsin(u)$ . So, the integral becomes $\arcsin(u) + C$ .
Substitute back $u$ : Substitute back $u = x + 7$ to get the final answer. $\arcsin(x + 7) + C.$
Match with options: Now, let's match our answer with the given options. $\newline$ The correct answer is $(B) \arcsin(x+7)+C$ .