What is the integral of sqrt(1+9x) ?

The integral of sqrt(1+9x) is 2/27 (1+9x)^{3/2}+C

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integral of sqrt(1+9x)

\int 1 + 9 x d x

\frac{2}{27} (1 + 9 x)^{\frac{3}{2}} + C

Solution steps

\int 1 + 9 x 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 = 1 + 9 x

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

Simplify

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

= \frac{2}{27} (1 + 9 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{27} (1 + 9 x)^{\frac{3}{2}} + C

x y^{^{''}} + 2 y^{^{'}} = 16 x^{2}

\int_{0}^{N} e^{(- s t)} e^{- 5 t} d t

2000 + 530 (ln (x))

\int_{1}^{16} t ln (t) d t

x tan (x)

What is the integral of sqrt(1+9x) ?
The integral of sqrt(1+9x) is 2/27 (1+9x)^{3/2}+C