First slide

Methods of integration

Question

$If \int \frac{(x - 1)^{2}}{x^{4} + x^{2} + 1} d x = f (x) + C, then the value of lim_{x \to \infty} f (x) is$

Moderate

A

$\frac{- π}{2 \sqrt{3}}$
B

$\frac{π}{2 \sqrt{3}}$
C

$\frac{π}{\sqrt{3}}$
D

$- \frac{2 π}{\sqrt{3}}$

Solution

$\begin{array}{l} \int \frac{(x - 1)^{2}}{x^{4} + x^{2} + 1} dx \\ = \int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} dx - \int \frac{2 x}{x^{4} + x^{2} + 1} dx \\ = \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} dx - \int \frac{2 x}{{(x^{2} + \frac{1}{2})}^{2} + \frac{3}{4}} dx \\ = \frac{1}{\sqrt{3}} \tan^{- 1} \frac{x - \frac{1}{x}}{\sqrt{3}} - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + C \\ = f (x) + C \\ ∴ lim_{x \to \infty} f (x) = \frac{1}{\sqrt{3}} \cdot \frac{π}{2} - \frac{2}{\sqrt{3}} \cdot \frac{π}{2} = \frac{- π}{2 \sqrt{3}} \end{array}$

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