Question

\[\int\frac{x}{3 x^4 - 18 x^2 + 11} dx\]

Answer 1

` ∫   {x  dx}/{3 x^4 - 18 x^2 + 11}`
\[\text{ let } x^2 = t\]
\[ \Rightarrow \text{ 2x dx }= dt\]
\[ \Rightarrow \text{ x dx }= \frac{dt}{2}\]
Now, ` ∫   {x  dx}/{3 x^4 - 18 x^2 + 11}`
\[ = \frac{1}{2}\int\frac{dt}{3 t^2 - 18t + 11}\]
\[ = \frac{1}{3 \times 2}\int\frac{dt}{t^2 - 6t + \frac{11}{3}}\]
\[ = \frac{1}{6}\int\frac{dt}{t^2 - 6t + 9 - 9 + \frac{11}{3}}\]
\[ = \frac{1}{6}\int\frac{dt}{\left( t - 3 \right)^2 - \frac{16}{3}}\]
\[ = \frac{1}{6}\int\frac{dt}{\left( t - 3 \right)^2 - \left( \frac{4}{\sqrt{3}} \right)^2}\]
\[ = \frac{1}{6} \times \frac{1}{2 \times \frac{4}{\sqrt{3}}} \text{ log  }\left| \frac{t - 3 - \frac{4}{\sqrt{3}}}{t - 3 + \frac{4}{\sqrt{3}}} \right| + C\]
\[ = \frac{\sqrt{3}}{48} \text{ log  }\left| \frac{x^2 - 3 - \frac{4}{\sqrt{3}}}{x^2 - 3 + \frac{4}{\sqrt{3}}} \right| + C\]