What is the integral of 1/(sqrt(a-bx)) ?

The integral of 1/(sqrt(a-bx)) is -(2sqrt(a-bx))/b+C

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integral of 1/(sqrt(a-bx))

\int \frac{1}{a - b x} d x

- \frac{2 a - b x}{b} + C

Solution steps

\int \frac{1}{a - b x} d x

Apply u-substitution

: \int - \frac{1}{b u} d u

= \int - \frac{1}{b u} d u

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - \frac{1}{b} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = a - b x

= - \frac{1}{b} \cdot 2 (a - b x)^{\frac{1}{2}}

Simplify

- \frac{1}{b} \cdot 2 (a - b x)^{\frac{1}{2}} : - \frac{2 a - b x}{b}

= - \frac{2 a - b x}{b}

Add a constant to the solution

= - \frac{2 a - b x}{b} + C

y = x^{2} - 1

x y^{^{'}} + y + y ln (x y) = 0

r \to 0 lim (\frac{1}{r ^{2}})

\int 6 e^{- 0.4 x} d x

\int \frac{6}{2} d x

What is the integral of 1/(sqrt(a-bx)) ?
The integral of 1/(sqrt(a-bx)) is -(2sqrt(a-bx))/b+C