How do I find this integral?

Question

2x/x^2+4*dx

Answer 1

I = ∫ (2x) / (x ² + 4) dx
Top line is derivative of bottom line:-
I = log (x ² + 4) + C

Answer 2

I am guessing you mean:

[2x/(x^2+4)]*dx

If so use substitution, let u = (x^2+4), then du = 2xdx

Then you have the integral of du/u which is ln(u), plug u = (x^2+4) back into it to get the answer:

ln(x^2 + 4)

Answer 3

Int 2x/(x^2 + 4) dx

Let u = x^2
du = 2x dx
Int = Int du/(u+1)
= ln(u +1) + C
= ln(x^2 + 1) + C

Answer 4

Use the substitution rule, using the denominator.

u = x^2 +4
du = 2x dx
Notice this is similar to the numerator, so just plug in du for numerator and u for denominator. It will look like this:

Integral ( du/u), which gives ln abs(u) abs = "absolute value"
Plug in u, which is x^2 +4 to get:

ln (x^2 +4) +C (with the inner equation being in the abs sign).

Answer 5

int [ 2x/(x^2 + 4) dx]

make a substitution,
let, u = x^2 + 4
du = 2x dx

int [ du/u ]
= ln(u) + c
= ln(x^2 + 4) + c

Answer 6

first ask...is the derivative of the bottom in the top? it is...so you can use:
(top ln|bottom|)/der. of bottom

which is
(2x ln|x^2|)/2x + 4x + C
=ln|x^2| + 4x + C