∫t4dt1−t10dt=∫t4dt1−(t5)2t5=u⇒5t4dt=du Use ∫1a2−x2dx=12alog⁡a+xa−x+c=15∫du12−u2=15⋅12(1)log⁡1+u1−u+c=110log1+t51−t5+c

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$\int \frac{t^{4} d t}{1 - t^{10}} i s$

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$\frac{1}{10} \log |\frac{1 - t^{5}}{1 + t^{5}}| + c$

$\frac{1}{5} \log |\frac{1 + t^{5}}{1 - t^{5}}| + c$

$\frac{- 1}{10} \log |\frac{1 - t^{5}}{1 + t^{5}}| + c$

$\frac{1}{10} \log |\frac{1 + t^{5}}{1 - t^{5}}| + c$

answer is D.

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Detailed Solution

$\int \frac{t^{4} d t}{1 - t^{10}} d t = \int \frac{t^{4} d t}{1 - {(t^{5})}^{2}}$

$t^{5} = u \Rightarrow 5 t^{4} d t = d u$

$Use \int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} \log |\frac{a + x}{a - x}| + c = \frac{1}{5} \int \frac{d u}{1^{2} - u^{2}}$

$= \frac{1}{5} \cdot \frac{1}{2 (1)} \log |\frac{1 + u}{1 - u}| + c$

$= \frac{1}{10} \log |\frac{1 + t^{5}}{1 - t^{5}}| + c$

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