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Given f(x)= 6x/ (3x^2 + 4)
We need to find the integral of f(x).
Let us assume that y= 3x^2 + 4
==> dy = 6x dx
Let us substitute:
intg f(x) = intg ( 6x/ (3x^2 + 4) dx
= intg ( dy / y)
= ln y + C
==> Now we will substitute with y= 3x^2 + 4
==> intg f(x) = ln ( 3x^2 + 4) + C
It is given that f(x) = 6x / (3x^3 + 4)
Let t = 3x^2 + 4
dt / dx = 6x
=> dt = 6x dx
Int [6x / (3x^3 + 4) dx]
=> Int [dt / t]
=> ln t + C
replace t with 3x^2 + 4
=> ln (3x^2 + 4) + C
Therefore the integral of f(x) = 6x / (3x^2 + 4) is ln (3x^2 + 4) + C.
To find the integral 6x/(3x^2+4)
We put t = 3x^2+4.
We differentiate t = 3x^2+4.
di = 6xdx.
Therefore Int f(x) dx = Int (6xdx/(3x^2+4)
Int (6xdx/(3x^2+4) = Int dt//t
Int (6xdx/(3x^2+4) = lnt +C...(1)
Now we put t= 3x^2+4 in (1).
Therefore (6xdx/(3x^2+4) = ln(3x^2+4) + C, where C is the constant of integration.
We notice that the numerator is the derivative of denominator. So, we'll note the denominator by u:
3x^2 + 4 = u
3x^2 = u - 4
x^2 = (u-4)/3
x = sqrt[(u-4)/3]
We'll differentiate both sides:
6xdx = du
We'll re-write the integral in u:
Int 6xdx / (3x^2 + 4) = Int du/u
Int du/u = ln |u| + C
Int 6xdx / (3x^2 + 4) = ln (3x^2 + 4) + C
Since 3x^2 + 4 > 0, we will not take the absolute value of 3x^2 + 4.
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