What is the integral of 1/(sqrt(25-x)) ?

The integral of 1/(sqrt(25-x)) is -2sqrt(25-x)+C

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integral of 1/(sqrt(25-x))

\int \frac{1}{25 - x} d x

- 2 25 - x + C

Solution steps

\int \frac{1}{25 - x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 25 - x

= - 2 (25 - x)^{\frac{1}{2}}

Apply radical rule:

= - 2 25 - x

Add a constant to the solution

= - 2 25 - x + C

n = 1 \sum \infty (2 x)^{n}

\frac{\partial}{\partial x} (y^{2} - 3 x^{3})

4 + t^{6}

f (x) = - 5 x

y^{^{'''}} - 7 y^{^{'}} + 6 y = e^{3 x}

What is the integral of 1/(sqrt(25-x)) ?
The integral of 1/(sqrt(25-x)) is -2sqrt(25-x)+C