∫ ( 3 / (square root of (3x + 1))) dx =?

Question

Hmm i'm not sure if i'm happy on how i've typed this- square root signs can be a pain here i guess...
Thanks

Answer 1

first off, where'd you find the integral sign?

Second, if you're looking for the antiderivative, you have to think of what would give a derivative of 3/(sqrt(3x+1)). Because you have a sqrt(3x+1) in the denominator, and a constant on top, recall that d/dx (k*f(x)) =k *d/dx (f(x)).

The derivative of sqrt(3x+1) would be 1/(2sqrt(3x+1)) d/dx (3x +1). and d/dx (3x+1) = 3, explaining the 3 in front. Thus, the antiderivative of 3/sqrt(3x+1) = 2sqrt(3x+1) + C to compensate for the 1/2 that will be multiplied to the derivative

Answer 2

Another way.. Note: Int means integrate!

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

This is in the form of ( u * du/dx) (Ignoring the power -1/2 for now) meaning if you differentate 3x+1 you get 3. So if its in this kind of form the answer will be "Increase the u terms power by 1 and divide by the same power.

u term (with power) is (3x+1)^(-1/2)

Ans (3x+1)^(-1/2 + 1)/(-1/2 + 1)
= 2(3x+1)^1/2

Hope this helps!

Answer 3

∫ ( 3 / (square root of (3x + 1))) dx

use u = 3x + 1 and du = 3 dx
∫ 1 / sqrt(u) du
∫ u^(-1/2) du
= 2u^(1/2)
= 2(3x+1)^(1/2)

Answer 4

let u =3x+1, then du = 3 dx

So integral(3/sqrt(3x+1) dx) = integral(du/sqrt(u)) = 2*sqrt(u) = 2*sqrt(3x+1)