What is the integral of (10x-2)/(5x^2-2x) ?

The integral of (10x-2)/(5x^2-2x) is ln|5x-2|+ln|x|+C

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integral of (10x-2)/(5x^2-2x)

\int \frac{10 x - 2}{5 x ^{2} - 2 x} d x

ln ∣ 5 x - 2 ∣ + ln ∣ x ∣ + C

Solution steps

\int \frac{10 x - 2}{5 x ^{2} - 2 x} d x

Expand

\frac{10 x - 2}{5 x ^{2} - 2 x} : \frac{10 x}{5 x ^{2} - 2 x} - \frac{2}{5 x ^{2} - 2 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{10 x}{5 x ^{2} - 2 x} d x - \int \frac{2}{5 x ^{2} - 2 x} d x

\int \frac{10 x}{5 x ^{2} - 2 x} d x = 2 ln ∣ 5 x - 2 ∣

\int \frac{2}{5 x ^{2} - 2 x} d x = - ln ∣ x ∣ + ln ∣ 5 x - 2 ∣

= 2 ln ∣ 5 x - 2 ∣ - (- ln ∣ x ∣ + ln ∣ 5 x - 2 ∣)

Simplify

= ln ∣ 5 x - 2 ∣ + ln ∣ x ∣

Add a constant to the solution

= ln ∣ 5 x - 2 ∣ + ln ∣ x ∣ + C

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

x \to 0 - lim (csc (x))

y^{^{''}} - 25 y = 0, y (2) = - 2, y^{^{'}} (- 2) = 2

y^{^{'}} = y cos (t)

x cos (4 x)

What is the integral of (10x-2)/(5x^2-2x) ?
The integral of (10x-2)/(5x^2-2x) is ln|5x-2|+ln|x|+C