Question

Ayuda, por favor!
Integral que conduce a función trigonométrica inversa.
$\int\limits{ \frac{2 x^{3} }{2 x^{2} -4x+3} } \, dx$

Answer 1

$\displaystyle I= \int\limits{ \frac{2 x^{3} }{2 x^{2} -4x+3} } \, dx \\ \\ \\ \texttt{Primero hay que dividir:}\\ \\ \frac{2 x^{3} }{2 x^{2} -4x+3} =x + 2+\frac{5x - 6}{2x^2 - 4x + 3}\\ \\ \\ I=\int x+2 ~dx+\int \frac{5x - 6}{2x^2 - 4x + 3}~dx\\ \\ \\ I=x^2/2+2x+\int \frac{5x - 6}{2x^2 - 4x + 3}~dx$


$\displaystyle \texttt{Observamos que } d(2x^2 - 4x + 3)=(4x-4)~dx\texttt{ entonces llevamos}\\ \texttt{a }5x-6\texttt{ a esa forma }5x-6=\frac{5}{4}(4x-4)-1\\ \\ \\ I=x^2/2+2x+\frac{5}{4}\int \frac{4x - 4}{2x^2 - 4x + 3}~dx-\int \frac{1}{2x^2 - 4x + 3}~dx\\ \\ \\ I=x^2/2+2x+\frac{5}{4}\int \frac{d(2x^2 - 4x + 3)}{2x^2 - 4x + 3}-\int \frac{1}{2x^2 - 4x + 3}~dx$


$\displaystyle I=x^2/2+2x+\frac{5}{4}\ln\left|2x^2 - 4x + 3\right|-\int \frac{1}{2x^2 - 4x + 3}~dx\\ \\ \\ I=x^2/2+2x+\frac{5}{4}\ln\left|2x^2 - 4x + 3\right|-\int \frac{1}{2(x-1)^2+1}~dx\\ \\ \\ I=x^2/2+2x+\frac{5}{4}\ln\left|2x^2 - 4x + 3\right|-\frac{1}{2}\int \frac{1}{(x-1)^2+\frac{1}{2}}~dx\\ \\ \\ \boxed{I=x^2/2+2x+\frac{5}{4}\ln\left|2x^2 - 4x + 3\right|-\frac{\sqrt{2}}{2}\arctan\left[\sqrt{2}(x-1)\right]+C}$