What is the integral of 1/(x(1-0.0004x)) ?

The integral of 1/(x(1-0.0004x)) is ln|x|-ln|0.0004x-1|+C

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integral of 1/(x(1-0.0004x))

\int \frac{1}{x ( 1 - 0.0004 x )} d x

ln ∣ x ∣ - ln ∣ 0.0004 x - 1 ∣ + C

Solution steps

\int \frac{1}{x ( 1 - 0.0004 x )} d x

Take the partial fraction of

\frac{1}{x ( 1 - 0.0004 x )} : \frac{1}{x} - \frac{0.0004}{0.0004 x - 1}

= \int \frac{1}{x} - \frac{0.0004}{0.0004 x - 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x} d x - \int \frac{0.0004}{0.0004 x - 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{0.0004}{0.0004 x - 1} d x = ln ∣ 0.0004 x - 1 ∣

= ln ∣ x ∣ - ln ∣ 0.0004 x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ 0.0004 x - 1 ∣ + C

\frac{d}{d x} (a cos (2 x) + b sin (2 x))

\frac{d}{d x} (a sin (x))

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 4

\int tan (2 x se) c^{2} 2 x d x

y = 3 x - 2 x^{2}, (- 3, - 27)

What is the integral of 1/(x(1-0.0004x)) ?
The integral of 1/(x(1-0.0004x)) is ln|x|-ln|0.0004x-1|+C