What is the integral of sqrt(3t+1) ?

The integral of sqrt(3t+1) is 2/9 (3t+1)^{3/2}+C

integral of sqrt(3t+1)

\int 3 t + 1 d t

\frac{2}{9} (3 t + 1)^{\frac{3}{2}} + C

Solution steps

\int 3 t + 1 d t

Apply u-substitution

= \int \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u d u

Apply the Power Rule

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 t + 1

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

Simplify

\frac{1}{3} \cdot \frac{2}{3} (3 t + 1)^{\frac{3}{2}} : \frac{2}{9} (3 t + 1)^{\frac{3}{2}}

= \frac{2}{9} (3 t + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (3 t + 1)^{\frac{3}{2}} + C

\int \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x

\int 68 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

\int \frac{1}{4 t ^{3}} d t

\int \frac{8}{x + 2} d x

\int x^{2} arcsec (x) d x

What is the integral of sqrt(3t+1) ?
The integral of sqrt(3t+1) is 2/9 (3t+1)^{3/2}+C