How would you find an antiderivative for this?

Question

Given g'(x) = 3x^2 + 5x + 4 / sqrt x (read as three x squared plus 5x plus 4 all over the square root of x) , which of the following is an antiderivative? I know that one method of doing this is to find the derivative of each of the choices and see if they match up with the g'(x), but i can't figure out how to do this. Please help.

A) sqrtx(3x^2 + 5x + 4) +C,
B) sqrtx(3/5x^2 + 5/3x + 4) +C,
C) 2sqrtx(3/5x^2 + 5/3x - 4) +C,
D) 2sqrtx(3x^2 + 5x + 4) +C,
E) 2sqrtx(3x^2 + 5x - 4) +C,
F) 2sqrtx(3/5x^2 + 5/3x + 4) +C,

In the answer choices, read the frations as 3 fifths and five 3rds instead of 3 over 5x and 5 over 3x. I wasn't sure if those were the same thing or not.

Answer 1

1) g'(x) = (3x^2+5x+4)/sqrt(x)
Recall that sqrt(x) = x^(1/2) or 1/sqrt(x) = x^(-1/2) therefore
multiply and add exponents..
2) g'(x) = 3x^(3/2)+5x^(1/2)+4*x^(-1/2)
Of course the anti derivative of x^n is (x^(n+1))/n+1 thus
3) g(x) = 6/5*x^(5/2)+10/3x^(3/2)
+8*x^(1/2) + C
factor out 2*x^(1/2)
4) g(x) = 2*x^(1/2)*(3/5x^2+5/3x+4) + C
of course 2*x^(1/2) = 2sqrtx so
5) g(x) = 2sqrtx*(3/5x^2+5/3x+4) + C

(F) is shown as the correct answer

Note: Dr. Pratt read the problem as you stated it g'(x) = 3x^2+5x+4 / sqrt(x). I gather from your explanation you meant to state it as (3x^2+5x+4)/sqrt(x). Without the parentheses the order of operations and the consequently the expressions are quite different.

Answer 2

Finding the derivative of each choice is going to take far too long and thats not what you want on a test or exam. To antidifferentiate:

1 Put the equation in the form that you want. In this case i would bring the square root x to the top making it to the power of negative a half

3x^2 + 5x + 4 x^(-1/2)

2 Use the basic antidifferentiation rule for this question, increase the power by one and divide by the new power, hence

3x^3/3 + 5x^2/2 + 4 x^(1/2)/ 1/2 / 4 + c

3 Cancel out and simplify

x^3 + 5x^2 + 8 x^(1/2) + c


Dr Pratt

Answer 3

you may not get an indefinite necessary for this, yet you have some alternatives, at the beginning show sin (x^2) as an endless series (i.e use a Taylor growth) and combine term by making use of term. you may hire a polar necessary like the case of ? e^(-x^2) dx to get the required from -infinity to infinity or 0 to infinity. If of course f'(x) = 2x sin (x^2) permit t = x^2 dt/dx = 2x then it turns into ? sin t dt = -cos t +c = -cos (x^2) + c the place c is an arbitrary consistent of integration.