Integrate: u 2 1 - u 6 ? d u

Question

Integrate: $\int \frac{u^{2}}{1 - u^{6}} d u$

Smart Achievers · Accepted Answer

(1)

Ans. $\frac{1}{6} \log$ $| \frac{1 + u^{3}}{1 - u^{3}} |$ $+ C$

Explanation

$The given integral is$ $\int \frac{u^{2}}{1 - u^{6}} d u$

$= \int \frac{u^{2}}{1 - {(u^{3})}^{2}} d u$

$Substitute z for u^{3} and differentiate w.r.t. u .$

$u^{3} = z$

$\frac{d}{d u} (u^{3}) = \frac{d z}{d u}$

$3 u^{2} = \frac{d z}{d u}$

$3 u^{2} d u = d z$

$\int \frac{u^{2}}{1 - {(u^{3})}^{2}} d u = \frac{1}{3} \int \frac{d z}{1 - z^{2}}$

$= \frac{1}{3} [\frac{1}{2} \log | \frac{1 + z}{1 - z} |] + C$

$= \frac{1}{3} [\frac{1}{2} \log | \frac{1 + u^{3}}{1 - u^{3}} |] + C$ $(∵ z = u^{3})$

$= \frac{1}{6} \log$ $| \frac{1 + u^{3}}{1 - u^{3}} |$ $+ C$

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