What is the integral of sqrt(1-4x) ?

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

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integral of sqrt(1-4x)

\int 1 - 4 x d x

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

Solution steps

\int 1 - 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 = 1 - 4 x

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

Simplify

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

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

Add a constant to the solution

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

cot^{2} (sin (θ))

y = cos (sin (tan (π x)))

\frac{d y}{d x} = y^{(\frac{1}{3})}

\int_{0}^{7} 7 - x d x

\int \frac{1}{x ^{2} + 2 x + 65} d x

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