Find the General Anti-derivative (Calculus I)

[treehau5]

Homework Statement

given f(x) = [x^3+5sqrt(x)]/x^2, find the anti-derivative

Homework Equations

The Attempt at a Solution

Hi I have attempted to solve this by re-writing the equation as a sum of two fractions:

x^3/x^2 + 5sqrt(x)/x^2, simplying gives = x + 5/x^3/2

I then apply the principle anti-differentiation rules, and I come out with:

x^2/2 + 5x^-1/2 = x^2/2 + 10/sqrt(x) + C

This answer is coming back in WebWork as wrong. I also checked my answer on wolframalpha, and got the same result.

What am I doing wrong? (If anything)

 

[DryRun]

x^3/x^2 + 5sqrt(x)/x^2, simplying gives = x + 5/x^3/2

Try re-writing it: x + \frac{5}{x^{3/2}}=x+5x^{-3/2}Then integrate.

 

[treehau5]

I am not quite at intergration yet, this is the last section of my Calculus I course, Calculus II is intergration. So for now our anti-d's are pretty simple and just follow some basic rules.

 

[treehau5]

Ok so
x + 5x^-3/2 following the rule: x(n+1)/(n+1) gives

x¹⁺¹ / (1+1) + 5x^{-3/2 + 2/2} / -(1/2) =

x²/2 + 5x^-1/2/ -(1/2 ) = -10x^-1/2 or finally

x² / 2 + 10 / sqrt(x)

I am still doing it wrong?

 

[treehau5]

ok yes I am doing it wrong, its - 10 / sqrt(x).

Thank you very much.

Story of my life, I always miss a sign.

 

[Mark44]

I am not quite at intergration yet, this is the last section of my Calculus I course, Calculus II is intergration. So for now our anti-d's are pretty simple and just follow some basic rules.

Integration is finding the antiderivative.

Note that there is no such word in English as inter[/color]gration.

 

[treehau5]

Integration is finding the antiderivative.

Note that there is no such word in English as inter[/color]gration.

Thank you Mark for the insight. And you are right, there is no such word as intergration.