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∫t4dt1−t10is

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a

110log1−t51+t5+c

b

15log1+t51−t5+c

c

−110log1−t51+t5+c

d

110log1+t51−t5+c

answer is D.

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Detailed Solution

∫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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