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Find 
int(1)/(sqrt(-x^(2)-4x+21))dx.
Choose 1 answer:
(A) 
(1)/(5)arctan((x+2)/(5))+C
(B) 
(1)/(5)arcsin((x+2)/(5))+C
(c) 
arctan((x+2)/(5))+C
(D) 
arcsin((x+2)/(5))+C

Find 1x24x+21dx \int \frac{1}{\sqrt{-x^{2}-4 x+21}} d x .\newlineChoose 11 answer:\newline(A) 15arctan(x+25)+C \frac{1}{5} \arctan \left(\frac{x+2}{5}\right)+C \newline(B) 15arcsin(x+25)+C \frac{1}{5} \arcsin \left(\frac{x+2}{5}\right)+C \newline(c) arctan(x+25)+C \arctan \left(\frac{x+2}{5}\right)+C \newline(D) arcsin(x+25)+C \arcsin \left(\frac{x+2}{5}\right)+C

Full solution

Q. Find 1x24x+21dx \int \frac{1}{\sqrt{-x^{2}-4 x+21}} d x .\newlineChoose 11 answer:\newline(A) 15arctan(x+25)+C \frac{1}{5} \arctan \left(\frac{x+2}{5}\right)+C \newline(B) 15arcsin(x+25)+C \frac{1}{5} \arcsin \left(\frac{x+2}{5}\right)+C \newline(c) arctan(x+25)+C \arctan \left(\frac{x+2}{5}\right)+C \newline(D) arcsin(x+25)+C \arcsin \left(\frac{x+2}{5}\right)+C
  1. Recognize integral form: Next, recognize that the integral is in the form of an arcsin function.\newlineThe general form is 1a2u2du=1aarcsin(ua)+C\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \frac{1}{a}\arcsin(\frac{u}{a}) + C.\newlineHere, a=5a = 5 and u=x+2u = x + 2.\newlineSo, the integral becomes (15)arcsin(x+25)+C(\frac{1}{5})\arcsin(\frac{x + 2}{5}) + C.
  2. Apply arcsin function: Now, check the answer choices to see which one matches our result.\newlineThe correct answer is (B) (15)arcsin(x+25)+C(\frac{1}{5})\arcsin(\frac{x + 2}{5}) + C.

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