Integrating x/(ax^3 + bx -c ): Solutions & Steps Explained

[grey]

hi,

can anyone tell me how the following will be integrated:
(where all letters except 'x' are constants)

x/(ax^3 + bx -c )

i tried to simplify the integral using partial fractions, but ended up with:

p/(x-q) + (rx +s)/(tx^2 + ux + v)
obviously, the first term is trivial, but how to integrate the second term. (the quadratic expression in the denominator does not have real roots)

 

[g_edgar]

(rx +s)/(tx^2 + ux + v)

If the numerator is a constant times the derivative of the denominator, great, make a substitution.

Otherwise, complete the square in the denominator ... surely every calculus textbook has this method?

 

[HallsofIvy]

hi,

can anyone tell me how the following will be integrated:
(where all letters except 'x' are constants)

x/(ax^3 + bx -c )

i tried to simplify the integral using partial fractions, but ended up with:

p/(x-q) + (rx +s)/(tx^2 + ux + v)
obviously, the first term is trivial, but how to integrate the second term. (the quadratic expression in the denominator does not have real roots)

If a quadratic expression such as $tx^2+ ux+ v$ cannot be factored, then you can complete the square to get something like $t(x-a)^2+ b$ where b is positive.
Then $(rx+ x)/(t(x-a)^2+ b)$ can be written as $(r(x-a)+ ra+ a)/(t(x-a)^2+ b)$.
The first term, $r(x-a)dx/(t(x-a)^2+ b)$ can be integrated with the substitution u= (x-a)^2 so that (1/2)du= (x-a)dt and the integrand becomes (r/(2t))du/u. The second term, (ra+a)dx/(t(x-a)^2+ b)= 1/(b(ra+a)) dx/((t/b)(x-a)^2+ 1) can be integrated with the substitution $u= \sqrt{t}{b}(x-a)$ and the fact that $\int du/(u^2+1)= arctan(u)+ C$.

 

[grey]

thanks guys!

definite help! thanks!