What is the integral of (0.8x+1/(60^2))^{-1/2} ?

The integral of (0.8x+1/(60^2))^{-1/2} is 2.5(0.8x+0.00027…)^{1/2}+C

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integral of (0.8x+1/(60^2))^{-1/2}

\int (0.8 x + \frac{1}{6 0 ^{2}})^{- \frac{1}{2}} d x

2.5 (0.8 x + 0.00027 \dots)^{\frac{1}{2}} + C

Solution steps

\int (0.8 x + \frac{1}{6 0 ^{2}})^{- \frac{1}{2}} d x

Apply u-substitution

: \int \frac{1}{0.8 u ^{\frac{1}{2}}} d u

= \int \frac{1}{0.8 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}{0.8} \cdot \int \frac{1}{u ^{\frac{1}{2}}} d u

Simplify

= 1.25 \cdot \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

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

= 1.25 \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 0.8 x + \frac{1}{6 0 ^{2}}

= 1.25 \cdot 2 (0.8 x + \frac{1}{6 0 ^{2}})^{\frac{1}{2}}

Simplify

= 2.5 (0.8 x + 0.00027 \dots)^{\frac{1}{2}}

Add a constant to the solution

= 2.5 (0.8 x + 0.00027 \dots)^{\frac{1}{2}} + C

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What is the integral of (0.8x+1/(60^2))^{-1/2} ?
The integral of (0.8x+1/(60^2))^{-1/2} is 2.5(0.8x+0.00027…)^{1/2}+C