What is the integral of (0.001)/((0.001x+1)^2) ?

The integral of (0.001)/((0.001x+1)^2) is -1/(0.001x+1)+C

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integral of (0.001)/((0.001x+1)^2)

integral

\frac{0.001}{( 0.001 x + 1 ) ^{2}}

- \frac{1}{0.001 x + 1} + C

Solution steps

\int \frac{0.001}{( 0.001 x + 1 ) ^{2}} d x

Take the constant out:

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

= 0.001 \cdot \int \frac{1}{( 0.001 x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{0.001 u ^{2}} d u

= 0.001 \cdot \int \frac{1}{0.001 u ^{2}} d u

Take the constant out:

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

= 0.001 \cdot \frac{1}{0.001} \cdot \int \frac{1}{u ^{2}} d u

Simplify

= 0.001 \cdot 1000 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 0.001 \cdot 1000 (- \frac{1}{u})

Substitute back

u = 0.001 x + 1

= 0.001 \cdot 1000 (- \frac{1}{0.001 x + 1})

Simplify

0.001 \cdot 1000 (- \frac{1}{0.001 x + 1}) : - \frac{1}{0.001 x + 1}

= - \frac{1}{0.001 x + 1}

Add a constant to the solution

= - \frac{1}{0.001 x + 1} + C

x - 3

2 cos (x) - sin^{3} (x) cos^{3} (x)

\frac{1}{( 3 x - 1 )}

x^{2} (x^{3} + 2)^{4}

\frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

What is the integral of (0.001)/((0.001x+1)^2) ?
The integral of (0.001)/((0.001x+1)^2) is -1/(0.001x+1)+C