What is the integral of sqrt(9x+4) ?

The integral of sqrt(9x+4) is 2/27 (9x+4)^{3/2}+C

Popular >

integral of sqrt(9x+4)

\int 9 x + 4 d x

\frac{2}{27} (9 x + 4)^{\frac{3}{2}} + C

Solution steps

\int 9 x + 4 d x

Apply u-substitution

: \int \frac{u}{9} d u

= \int \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{9} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 x + 4

= \frac{1}{9} \cdot \frac{2}{3} (9 x + 4)^{\frac{3}{2}}

Simplify

\frac{1}{9} \cdot \frac{2}{3} (9 x + 4)^{\frac{3}{2}} : \frac{2}{27} (9 x + 4)^{\frac{3}{2}}

= \frac{2}{27} (9 x + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{27} (9 x + 4)^{\frac{3}{2}} + C

\frac{d}{d x} ((cos (x^{2}))^{2})

y^{^{''}} + 4 y = 0, y (\frac{π}{8}) = 0, y (\frac{π}{6}) = 1

\frac{d}{d x} (x^{x + 1})

(- 20, - 4), (- 12, - 10)

n = 1 \sum \infty \frac{2}{25} \cdot (\frac{1}{5})^{n}

What is the integral of sqrt(9x+4) ?
The integral of sqrt(9x+4) is 2/27 (9x+4)^{3/2}+C