What is the integral of 2/(sqrt(9-4x)) ?

The integral of 2/(sqrt(9-4x)) is -sqrt(9-4x)+C

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integral of 2/(sqrt(9-4x))

\int \frac{2}{9 - 4 x} d x

- 9 - 4 x + C

Solution steps

\int \frac{2}{9 - 4 x} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{9 - 4 x} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= 2 \cdot \int - \frac{1}{4 u} d u

Take the constant out:

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

= 2 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

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

Substitute back

u = 9 - 4 x

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

Simplify

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

= - 9 - 4 x

Add a constant to the solution

= - 9 - 4 x + C

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What is the integral of 2/(sqrt(9-4x)) ?
The integral of 2/(sqrt(9-4x)) is -sqrt(9-4x)+C