What is the integral of-(4000)/((1+0.4x)^2) ?

The integral of-(4000)/((1+0.4x)^2) is (10000)/(1+0.4x)+C

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integral of-(4000)/((1+0.4x)^2)

\int - \frac{4000}{( 1 + 0.4 x ) ^{2}} d x

\frac{10000}{1 + 0.4 x} + C

Solution steps

\int - \frac{4000}{( 1 + 0.4 x ) ^{2}} d x

Take the constant out:

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

= - 4000 \cdot \int \frac{1}{( 1 + 0.4 x ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{0.4 u ^{2}} d u

= - 4000 \cdot \int \frac{1}{0.4 u ^{2}} d u

Take the constant out:

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

= - 4000 \cdot \frac{1}{0.4} \cdot \int \frac{1}{u ^{2}} d u

Simplify

= - 4000 \cdot 2.5 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - 4000 \cdot 2.5 (- \frac{1}{u})

Substitute back

u = 1 + 0.4 x

= - 4000 \cdot 2.5 (- \frac{1}{1 + 0.4 x})

Simplify

- 4000 \cdot 2.5 (- \frac{1}{1 + 0.4 x}) : \frac{10000}{1 + 0.4 x}

= \frac{10000}{1 + 0.4 x}

Add a constant to the solution

= \frac{10000}{1 + 0.4 x} + C

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What is the integral of-(4000)/((1+0.4x)^2) ?
The integral of-(4000)/((1+0.4x)^2) is (10000)/(1+0.4x)+C