# Thread: antidifferentiation problem. is this correct?

1. ## antidifferentiation problem. is this correct?

For the rule integral sign 1/x dx = ln lxl . does it only apply for 1/x? I mean does integral sign 1/2x dx = ln l2xl ?

Also
The problem is:

integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

Is this correct:

integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

= -3 x -e^(-x) + 2 x (1/2) x^2 - (1/2)(1/0.5)(e^(.5x)) + C
=3(e^(-x))+x^2-(e^(.5x)) + C

2. Originally Posted by fabxx
For the rule integral sign 1/x dx = ln lxl . does it only apply for 1/x? I mean does integral sign 1/2x dx = ln l2xl ?

Also
The problem is:

integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

Is this correct:

integral sign [-3e^(-x) + 2x - (e^(0.5x))/2] dx

= -3 x -e^(-x) + 2 x (1/2) x^2 - (1/2)(1/0.5)(e^(.5x)) + C
=3(e^(-x))+x^2-(e^(.5x)) + C

$\int \frac{1}{2x} \, dx = \frac{1}{2}\int \frac{1}{x} \, dx = \frac{1}{2} \ln|x| + C = \ln{\sqrt{x}} + C
$

please do not use "x" as the symbol for multiplication, especially when using x as a variable. common form is to use the asterisk (*).

$\int -3e^{-x} + 2x - \frac{1}{2}e^{\frac{x}{2}} \, dx =
$

$3e^{-x} + x^2 - e^{\frac{x}{2}} + C$

remember you can check integrals by taking the derivative of your solution.