First slide

Methods of integration

Question

$\int \frac{x^{3} - x}{1 + x^{6}} d x is equal to$

Moderate

A

$\frac{1}{6} \log \frac{x^{4} - x^{2} + 1}{{(1 + x^{2})}^{2}} + C$
B

$\frac{1}{6} \tan^{- 1} \frac{{(x^{2} + 1)}^{2}}{2} + C$
C

$\log \frac{x^{4} - x^{2} + 1}{{(1 + x^{2})}^{2}} + C$
D

$\tan^{- 1} \frac{{(x^{2} + 1)}^{2}}{2} + C$

Solution

$\begin{array}{l} I = \frac{1}{2} \int \frac{t - 1}{1 + t^{3}} d t (where t = x^{2}) \\ = \frac{1}{2} \int (\frac{- 2}{3 (1 + t)} + \frac{1}{3} \frac{(2 t - 1)}{t^{2} - t + 1}) d t (After partial fractions) \\ = \frac{1}{6} \cdot \ln \frac{x^{4} - x^{2} + 1}{{(x^{2} + 1)}^{2}} + C \end{array}$

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