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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

= 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

= 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

in t e g r a l x - 3

in t e g r a l \frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

in t e g r a l 1. 3^{x}

in t e g r a l \frac{3 x}{( x ^{2} + 4 x + 6 )}

in t e g r a l \frac{( 2 x )}{( 4 + 3 x ) ^{2}}

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