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

The integral of sqrt(9-4x) is -1/6 (9-4x)^{3/2}+C

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integral of sqrt(9-4x)

\int 9 - 4 x d x

- \frac{1}{6} (9 - 4 x)^{\frac{3}{2}} + C

Solution steps

\int 9 - 4 x d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 - 4 x

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

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (9 - 4 x)^{\frac{3}{2}} : - \frac{1}{6} (9 - 4 x)^{\frac{3}{2}}

= - \frac{1}{6} (9 - 4 x)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (9 - 4 x)^{\frac{3}{2}} + C

arctan (cos (x))

\int \frac{2 x ^{2} - 24}{x ( x + 2 ) ^{3}} d x

\int \frac{( x - 2 ) ^{n}}{n} d x

\frac{d}{d x} (\frac{x}{2 x - 7})

\int_{9}^{25} \frac{x - 9}{x} d x

What is the integral of sqrt(9-4x) ?
The integral of sqrt(9-4x) is -1/6 (9-4x)^{3/2}+C