$\int \frac{d t}{(6 t - 1)}$ is equal to:

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

ln (6 t - 1) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{6} ln (6 t - 1) + C$
Formula : $\int \frac{d t}{a t + b} = \frac{1}{a} ln (a t + b) + C$

$\Rightarrow \int \frac{d t}{6 t - 1} = \frac{1}{6} ln (6 t - 1) + C$

Suggest Corrections

Similar questions

\int \frac{d t}{6 t - 1}

ω

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} + = 2 ω

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}

\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}

x = l o g_{a} (b c), y = l o g_{b} (c a), z = l o g_{c} (a b),

|\begin{matrix} - 1 & \cos C & \cos B \\ \cos C & - 1 & \cos A \\ \cos B & \cos A & - 1 \end{matrix}|

Join BYJU'S Learning Program