Assume, u = 6t – 1 dudt=6 ⇒ nbsp; du6=dt I=∫dt6t 1=16∫duu nbsp;=16 ln⁡ |u|+C =16 ln(6t 1)+C ...

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Antiderivative

Integrate:∫dt...

Question

Integrate: $\int \frac{d t}{6 t - 1}$

\frac{1}{6} l n (6 t - 1) + C

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\frac{1}{12} l n (6 t - 1) + C

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\frac{1}{6} l n (12 t - 1) + C

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\frac{6}{1} l n (6 t - 1) + C

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Solution

The correct option is A $\frac{1}{6} l n (6 t - 1) + C$
Assume,
$u = 6 t - 1$
$\frac{d u}{d t} = 6 \Rightarrow \frac{d u}{6} = d t$
$I = \int \frac{d t}{6 t - 1} = \frac{1}{6} \int \frac{d u}{u}$
$= \frac{1}{6} l n | u | + C$
$= \frac{1}{6} l n (6 t - 1) + C$

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