What is the integral of 2000(1+0.2x)^{-3/2} ?

The integral of 2000(1+0.2x)^{-3/2} is -(20000)/((1+0.2x)^{1/2)}+C

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integral of 2000(1+0.2x)^{-3/2}

\int 2000 (1 + 0.2 x)^{- \frac{3}{2}} d x

- \frac{20000}{( 1 + 0.2 x ) ^{\frac{1}{2}}} + C

Solution steps

\int 2000 (1 + 0.2 x)^{- \frac{3}{2}} d x

Take the constant out:

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

= 2000 \cdot \int (1 + 0.2 x)^{- \frac{3}{2}} d x

Apply u-substitution

: \int \frac{1}{0.2 u ^{\frac{3}{2}}} d u

= 2000 \cdot \int \frac{1}{0.2 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= 2000 \cdot \frac{1}{0.2} \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Simplify

= 2000 \cdot 5 \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

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

= 2000 \cdot 5 (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = 1 + 0.2 x

= 2000 \cdot 5 (- \frac{2}{( 1 + 0.2 x ) ^{\frac{1}{2}}})

Simplify

2000 \cdot 5 (- \frac{2}{( 1 + 0.2 x ) ^{\frac{1}{2}}}) : - \frac{20000}{( 1 + 0.2 x ) ^{\frac{1}{2}}}

= - \frac{20000}{( 1 + 0.2 x ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{20000}{( 1 + 0.2 x ) ^{\frac{1}{2}}} + C

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What is the integral of 2000(1+0.2x)^{-3/2} ?
The integral of 2000(1+0.2x)^{-3/2} is -(20000)/((1+0.2x)^{1/2)}+C