What is the integral of 1/(sqrt(6t+1)) ?

The integral of 1/(sqrt(6t+1)) is 1/3 sqrt(6t+1)+C

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integral of 1/(sqrt(6t+1))

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

\frac{1}{3} 6 t + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 t + 1

= \frac{1}{6} \cdot 2 (6 t + 1)^{\frac{1}{2}}

Simplify

\frac{1}{6} \cdot 2 (6 t + 1)^{\frac{1}{2}} : \frac{1}{3} 6 t + 1

= \frac{1}{3} 6 t + 1

Add a constant to the solution

= \frac{1}{3} 6 t + 1 + C

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What is the integral of 1/(sqrt(6t+1)) ?
The integral of 1/(sqrt(6t+1)) is 1/3 sqrt(6t+1)+C