First Order Differential Equations

[delta587]

So I was practicing for my test next week but I simply don't understand how the integral of the circled part happened. Can someone please tell me why integral 3x/2x^2+6 became 3/4 integral 4x/2x^2+6? And in the next line this was turned into ln. Why?

 

[Deleted member 4993]

So I was practicing for my test next week but I simply don't understand how the integral of the circled part happened. Can someone please tell me why integral 3x/2x^2+6 became 3/4 integral 4x/2x^2+6? And in the next line this was turned into ln. Why?

Is this your work you are reviewing - or somebody else's work?

\(\displaystyle \frac{3}{4}\left[ \frac{4x}{2x^2+6} \right]\) .............................. cancel 4 from the numerator and the denominator.

= \(\displaystyle \frac{3x}{2x^2+6}\)

Do you know that:

\(\displaystyle \int \frac{dx}{x} \) = ln(x) + C ........................this is how the "fraction" turns into "ln".

 

[Steven G]

The 3 was factored out. There needed to be a 4 in the numerator as the the derivative of the denomerator had a factor of 4 in it. So the 4 was put in the integrand and a 1/4 was put outside the integrand (since 4*1/4 = 1)

 

[Steven G]

You do realize that 4x/2x^2+6 is not the same as 4x/(2x^2+6).

4x/2x^2+6 =2x^3 + 6

 

[Shiloh]

To integrate 3x dx/(2x^2+ 6) let u= 2x^2+ 6. Then du= 4x dx so xdx= du/4.
3x dx/(2x^2+ 6)= (3/4) du/u. Integrating that, (3/4) ln(u)+ C= (3/4) ln(2x^2+ 6)+ C