What is the integral of sqrt(1+36x) ?

The integral of sqrt(1+36x) is 1/54 (1+36x)^{3/2}+C

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integral of sqrt(1+36x)

\int 1 + 36 x d x

\frac{1}{54} (1 + 36 x)^{\frac{3}{2}} + C

Solution steps

\int 1 + 36 x d x

Apply u-substitution

: \int \frac{u}{36} d u

= \int \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{36} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 36 x

= \frac{1}{36} \cdot \frac{2}{3} (1 + 36 x)^{\frac{3}{2}}

Simplify

\frac{1}{36} \cdot \frac{2}{3} (1 + 36 x)^{\frac{3}{2}} : \frac{1}{54} (1 + 36 x)^{\frac{3}{2}}

= \frac{1}{54} (1 + 36 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{54} (1 + 36 x)^{\frac{3}{2}} + C

\frac{d}{d t} (a e^{b t})

\frac{\partial}{\partial x} (e^{- t})

\frac{d}{d x} (\frac{1}{x} + \frac{1}{x ^{2}})

\frac{\partial}{\partial t} (sech^{2} (x - t))

What is the integral of sqrt(1+36x) ?
The integral of sqrt(1+36x) is 1/54 (1+36x)^{3/2}+C