Integration - Using a differential to evaluate the integral of a fraction

[alex325]

Hi there,

I'd really appreciate some help!

I have the following problem:



I understand that f'x = 9x^2 + 3, but I don't see how this helps.

I also understand that the integral of f'x/fx = ln|f(x)|+C and I'm assuming this is useful for this problem.

I know that if I take out a factor of 3 from f'(x) then I'm left with the same expression as is on the top of the fraction I'm asked to integrate - but if I take out a factor of 3 from the top of the fraction, don't I also have to then take a factor of 3 from the bottom of the fraction?

I hope I've made sense, and thanks in advance to anyone who might be able to help.

 

[Deleted member 4993]

Hi there,

I'd really appreciate some help!

I have the following problem:

View attachment 17746

I understand that f'x = 9x^2 + 3, but I don't see how this helps.

I also understand that the integral of f'x/fx = ln|f(x)|+C and I'm assuming this is useful for this problem.

I know that if I take out a factor of 3 from f'(x) then I'm left with the same expression as is on the top of the fraction I'm asked to integrate - but if I take out a factor of 3 from the top of the fraction, don't I also have to then take a factor of 3 from the bottom of the fraction?

I hope I've made sense, and thanks in advance to anyone who might be able to help.

I''ll do a similar but different problem:

$\displaystyle \int \frac{x^2 -3x - 1}{x^3 - \frac{9}{2} x^2 - 3x + 7} dx$

so f(x) = $\displaystyle x^3 - \frac{9}{2} x^2 - 3x + 7$

f'(x) = $\displaystyle 3x^2 - {9} x^2 - 3$

$\displaystyle \int \frac{x^2 -3x - 1}{x^3 - \frac{9}{2} x^2 - 3x + 7} dx$

= $\displaystyle \int \frac{\frac{1}{3}f'(x)}{f(x)} dx$

=$\displaystyle \frac{1}{3}\int \frac{f'(x)}{f(x)} dx$

= $\displaystyle \frac{1}{3}ln|f(x)| + C$

 

[alex325]

Oh yes I see exactly what to do now, thank you so much!

 

[topsquark]

Hi there,

I'd really appreciate some help!

I have the following problem:

View attachment 17746

I understand that f'x = 9x^2 + 3, but I don't see how this helps.

I also understand that the integral of f'x/fx = ln|f(x)|+C and I'm assuming this is useful for this problem.

I know that if I take out a factor of 3 from f'(x) then I'm left with the same expression as is on the top of the fraction I'm asked to integrate - but if I take out a factor of 3 from the top of the fraction, don't I also have to then take a factor of 3 from the bottom of the fraction?

I hope I've made sense, and thanks in advance to anyone who might be able to help.

$$f'(x) = 9x^2 + 3 = 3( 3x^2 + 1)$$. Hint, hint, hint....

-Dan