Find int(1)/(sqrt(-x^(2)-6x+40))dx. Choose 1 answer: (A) (1)/(14)arctan((x+3)/(7))+C (B) (1)/(14)arcsin((x+3)/(7))+C (c) arctan((x+3)/(7))+C () arcsin((x+3)/(7))+C

Recognize inverse trigonometric function: Next, we recognize that the integral is in the form of an inverse trigonometric function, specifically arcsin. The general form is int 1/sqrt(a^2 - u^2) du = (1/a)arcsin(u/a) + C. Here, a = 7 and u = x + 3. So the integral becomes (1/7)arcsin((x + 3)/7) + C.Substitute values: Finally, we check the answer choices to see which one matches our result. The correct answer is (B) (1/14)arcsin((x + 3)/7) + C. Wait, we made a mistake in the coefficient. It should be (1/7) not (1/14).

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Find
int(1)/(sqrt(-x^(2)-6x+40))dx.
Choose 1 answer:
(A)
(1)/(14)arctan((x+3)/(7))+C
(B)
(1)/(14)arcsin((x+3)/(7))+C
(c)
arctan((x+3)/(7))+C
()
arcsin((x+3)/(7))+C

Find $\int \frac{1}{\sqrt{-x^{2}-6 x+40}} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{14} \arctan \left(\frac{x+3}{7}\right)+C$ $\newline$ (B) $\frac{1}{14} \arcsin \left(\frac{x+3}{7}\right)+C$ $\newline$ (c) $\arctan \left(\frac{x+3}{7}\right)+C$ $\newline$ () $\arcsin \left(\frac{x+3}{7}\right)+C$

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Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{\sqrt{-x^{2}-6 x+40}} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{14} \arctan \left(\frac{x+3}{7}\right)+C

\newline

(B)

\frac{1}{14} \arcsin \left(\frac{x+3}{7}\right)+C

\newline

(c)

\arctan \left(\frac{x+3}{7}\right)+C

\newline

()

\arcsin \left(\frac{x+3}{7}\right)+C

Recognize inverse trigonometric function: Next, we recognize that the integral is in the form of an inverse trigonometric function, specifically $\arcsin$ . The general form is $\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \frac{1}{a}\arcsin\left(\frac{u}{a}\right) + C$ . Here, $a = 7$ and $u = x + 3$ . So the integral becomes $\frac{1}{7}\arcsin\left(\frac{x + 3}{7}\right) + C$ .
Substitute values: Finally, we check the answer choices to see which one matches our result. $\newline$ The correct answer is (B) $(\frac{1}{14})\arcsin(\frac{(x + 3)}{7}) + C$ . $\newline$ Wait, we made a mistake in the coefficient. It should be $(\frac{1}{7})$ not $(\frac{1}{14})$ .