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Answer:

$\frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{3}}}\right|+C$

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

$\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$

Solution:

$Let\: \: I=\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$

$=\int \frac{x}{3 (x^{2})^{2}-18 x^{2}+11} d x$

$Put\: \: x^{2}=t\Rightarrow 2x\, dx=dt\Rightarrow x\, dx=\frac{dt}{2}$

$Then\: \: I=\int \frac{1}{3t^{2}-18 t+11} \, \frac{dt}{2}$

\begin{aligned} &=\frac{1}{2} \int \frac{1}{3\left(t^{2}-6 t+\frac{11}{3}\right)} d t \\ &=\frac{1}{6} \int \frac{1}{t^{2}-2.3 . t+(3)^{2}-(3)^{2}+\frac{11}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-9+\frac{11}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\left(\frac{27-11}{3}\right)} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\frac{16}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} d t \end{aligned}

$Put\: \: t-3=u\Rightarrow dt=du$

$Then\: \: I =\frac{1}{6} \int \frac{1}{u^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} du$

\begin{aligned} &=\frac{1}{6} \cdot \frac{1}{2 \cdot \frac{4}{\sqrt{3}}} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{\sqrt{3}}{6 \times 8} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+C \\ &=\frac{\sqrt{3}}{48} \log \left|\frac{t-3-\frac{4}{\sqrt{3}}}{t-3+\frac{4}{\sqrt{3}}}\right|+C \quad[\because u=t-3] \end{aligned}

$=\frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{3}}}\right|+C\: \: \: \: \: \: \: [\because t=x^{2}]$

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