Evaluate the following integrals: x 3x^4-18x^2+11 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}}{3\text{x}^4-18\text{x}^2+11}\text{dx}$

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Answer

$\int\frac{\text{x dx}}{3\text{x}^4-18\text{x}^2+11}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x}\text{ dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{3\text{x}^4-18\text{x}^2+11}$
$=\frac{1}2{}\int\frac{\text{dt}}{3\text{t}^2-18\text{t}+11}$
$=\frac{1}{3\times2}\int\frac{\text{dt}}{\text{t}^2-6\text{t}+\frac{11}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2-6\text{t}+9-9+\frac{11}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{(\text{t}-3)^2-\frac{16}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{(\text{t}-3)^2-\Big(\frac{4}{\sqrt{3}}\Big)^2}$
$=\frac{1}{6}\times\frac{1}{2\times\frac{4}{\sqrt{3}}}\log\Bigg|\frac{\text{t}-3-\frac{4}{\sqrt{3}}}{\text{t}-3+\frac{4}{\sqrt{3}}}\Bigg|+\text{C}$
$=\frac{\sqrt{3}}{48}\log\Bigg|\frac{\text{x}^2-3-\frac{4}{\sqrt{3}}}{\text{x}^2-3+\frac{4}{\sqrt{3}}}\Bigg|+\text{C}$

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