Question

\[\int\frac{x^3 - 3x}{x^4 + 2 x^2 - 4}dx\]

Answer 1

\[I = \int\frac{x^3 - 3x}{x^4 + 2 x^2 - 4}dx\]

\[= \int\frac{x( x^2 - 3)}{x^4 + 2 x^2 - 4}dx\]

Let 

\[x^2 = t\] , or , 

\[2xdx = dt\]

\[\Rightarrow I = \frac{1}{2}\int\frac{(t - 3)}{t^2 + 2t - 4}dt\]
\[ = \frac{1}{4}\int\frac{2t - 6}{t^2 + 2t - 4}dt\]
\[ = \frac{1}{4}\int\frac{2t + 2 - 8}{t^2 + 2t - 4}dt\]
\[ = \frac{1}{4}\int\left( \frac{2t + 2}{t^2 + 2t - 4} - \frac{8}{t^2 + 2t - 4} \right)dt\]
\[ = \frac{1}{4}\left( \int\frac{2t + 2}{t^2 + 2t - 4}dt - \int\frac{8}{t^2 + 2t - 4}dt \right)\]

\[\Rightarrow I = \frac{1}{4}\left( I_1 + I_2 \right) . . . \left( i \right)\]

Now,

\[I_1 = \int\frac{2t + 2}{t^2 + 2t - 4} dt\]

\[t^2 + 2t - 4 = u\]

\[or, \left( 2t + 2 \right)dt = du\]
\[ \Rightarrow I_1 = \int\frac{1}{u} du = \text{ ln }\left| u \right| + c_1 \]
\[ \Rightarrow I_1 = \text{ ln }\left| t^2 + 2t - 4 \right| + c_1 \]
\[ \therefore I_1 = \text{ ln }\left| x^4 + 2 x^2 - 4 \right| + c_1\]

Now,

\[I_2 = \int\frac{- 8}{(t + 1 )^2 - 5}dt\]
\[ \Rightarrow I_2 = \int\frac{8}{(\sqrt{5} )^2 - (t + 1 )^2}dt\]
\[ \therefore I_2 = \frac{8}{2\sqrt{5}}\ln\left| \frac{\sqrt{5} + x^2 + 1}{\sqrt{5} - x^2 - 1} \right| + c_2\]

\[\text{ So, from }\left( i \right), \text{ we get}\]
\[I = \frac{1}{4}\left[ \text{ ln}\left| x^4 + 2 x^2 - 4 \right| + \frac{4}{\sqrt{5}}\text{ ln} \left| \frac{\sqrt{5} + x^2 + 1}{\sqrt{5} - x^2 - 1} \right| \right] + C\]
\[ \therefore I = \frac{1}{4}\text{ ln}\left| x^4 + 2 x^2 - 4 \right| + \frac{1}{\sqrt{5}}\text{ ln }\left| \frac{\sqrt{5} + x^2 + 1}{\sqrt{5} - x^2 - 1} \right| + C\]