How do I find an Antiderivitive with a square root in the denominator?

Question

I am working on a Calculus 2 homework problem and it wants me to evaluate the definate integral of (2x^2+6)/sqrt(x) between 4 and 9. How do I get the antiderivitave when there is a square root on the bottom?

Answer 1

The best way is to transform the expression before... and use the rule: anti-derivative of (x^n) is x^(n+1)/(n+1) for n different of -1
In this case it is possible:

sqrt(x) = x^(1/2) and 1/sqrt(x) = x^(-1/2)

So you get: (2x^2 + 6)/sqrt(x) = (2x^2 + 6)x^(-1/2)

Multiplying we get: 2x^2.x^(-1/2) + 6.x^(-1/2) = 2x^(3/2) +6.x^(-1/2)

Using the rule: 4x^(5/2)/5 + 12x^(1/2)

Answer 2

Split up the problem:
Integral of (2x^2+6)/sqrt(x) =
Integral of (2x^2/sqrt(x)) + Integral of (6/sqrt(x)) =
Integral of (2x^(3/2)) + Integral of (6x^(-1/2) =
(4/5)x^(5/2) + 12x^(1/2)

Now just evaluate it over the given range.

Answer 3

(2x^2+6)/sqrt(x) = 2 (x^2+3)/sqrt(x)

Now we can use variable substitution, like in this case, x = t^2

which leads to, 2(t^4 + 3)/t * 2t = 4 (t^4 + 3)
integrate it, and you have: 12t + 4/5 t^5, now since x = t^2

we have, 12sqrt(x) + 4/5 sqrt(x) x^2 = sqrt(x) (12 +4/5 x^2) = sqrt(x) 4/5 (15 + x^2)

now just apply barrow's rule.

Answer 4

Here's a hint: sqrt(x) = x^(1/2).

If you need futher help, let me know.