What is the integral of (1+5x)^{-1/2} ?

The integral of (1+5x)^{-1/2} is 2/5 (1+5x)^{1/2}+C

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integral of (1+5x)^{-1/2}

\int (1 + 5 x)^{- \frac{1}{2}} d x

\frac{2}{5} (1 + 5 x)^{\frac{1}{2}} + C

Solution steps

\int (1 + 5 x)^{- \frac{1}{2}} d x

Apply u-substitution

: \int \frac{1}{5 u ^{\frac{1}{2}}} d u

= \int \frac{1}{5 u ^{\frac{1}{2}}} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

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

= \frac{1}{5} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 + 5 x

= \frac{1}{5} \cdot 2 (1 + 5 x)^{\frac{1}{2}}

Simplify

\frac{1}{5} \cdot 2 (1 + 5 x)^{\frac{1}{2}} : \frac{2}{5} (1 + 5 x)^{\frac{1}{2}}

= \frac{2}{5} (1 + 5 x)^{\frac{1}{2}}

Add a constant to the solution

= \frac{2}{5} (1 + 5 x)^{\frac{1}{2}} + C

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What is the integral of (1+5x)^{-1/2} ?
The integral of (1+5x)^{-1/2} is 2/5 (1+5x)^{1/2}+C