Calculus:

[chosen5]

Integration by Substitution(reverse derivatives)
1:

x / (3x+2)^(1/2)

2:

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

3:

Sin4x

4:

(1+x^3)^1/2 * (x^2)

5:

(x^2+5)^3

 

[Jarconis]

you'll never use that in the real world....unless you grow up to be a calc teacher.

 

[Sir Savage]

If I do these problems for you, how does that help you?

Besides, "learning" calculus on a forum is not the best way to go about it.

 

[YellowJacket]

By God, I sucked at calculus, but damnit I can figure your BMR in seconds.

 

[Chemo]

Originally posted by chosen5
Integration by Substitution(reverse derivatives)
1:

x / (3x+2)^(1/2)

2:

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

3:

Sin4x

4:

(1+x^3)^1/2 * (x^2)

5:

(x^2+5)^3

1. [2*(3x-4) * (3x+2)^.5] / 27
2. [2*(5x-3)^.5] / 5
3. -45*cos(4x) / pi
4. [2*(x^3+1)^3/2] / 9
5. (x^7) / 7 + 3x^5 + 25x^3 + 125x

There's your answers (quick and dirty) but the actual work is up to you.

Chemo

 

[Jarconis]

hmm...chemo, can you do my taxes on here?