How to find the antiderivative of 4/(1+2x)^2?

[haiku11]

Homework Statement

I have to find the antiderivative of 4/(1+2x)²
New question: Find s(t)

2. Given variables and equations
a(t) = 4/(1+2x)²
v(0) = 0
s(0) = 5

The Attempt at a Solution

∫4(1+2x)^-2
= 4∫(1+2x)^-2
= 4[(1+2x)^-1/-1]
= -4/(1+2x)

But the correct answer has a -2 numerator.

New attempt: The integral of the first equation would be v(t) and it says v(0) is 0, but the equation says that it will be -2.

 

[HallsofIvy]

You are treating 1+ 2x as if it were a single variable- you can't do that. Use a "substitution" u= 1+ 2x. Then du= 2dx.

 

[hunt_mat]

You forgot to use the substitution u=2x, therefore du=2dx, and this will get you your -2 that you require.

 

[tiny-tim]

welcome to pf!

hi haiku11! welcome to pf!

another method is to use the chain rule …

d/dx = d/d(2x) times d(2x)/dx ​

 

[haiku11]

Oh ok that makes a lot more sense.

Thanks so much for your help.

Edit: I didn't see the rest of the question and now I'm even more confused. I edited the original post to show the entire question.

 

[haiku11]

Please Help!

 

[Ivan92]

I assume you have found the integral of your original function. Remember the ∫a(t) = v(t) + C. since you are given v(0)=0, you are supposed to solve for C. So now what you have to integrate v(t) because ∫v(t) = s(t) + Ct + D. You are given s(0)=0 so that will help you find D. Good luck :)

 

[tiny-tim]

lets see …

∫4(1+2x)^-2
= 4∫(1+2x)^-2
= 4[(1+2x)^-1/-1]
= -4/(1+2x)

But the correct answer has a -2 numerator.

first, you must write the dx at the end of each line …

∫ 4(1+2x)^-2 dx

= 4 ∫ (1+2x)^-2 dx

= 4 ∫ (1+2x)^-2 dx/d(2x) d(2x)

(you could put 2x = u here if you wanted)

= 4 ∫ (1+2x)^-2 1/2 d(2x)

= 2 [(1+2x)^-1/-1] + constant

= -2/(1+2x) + constant