What is the integral of 1/((5x^2+x)) ?

The integral of 1/((5x^2+x)) is ln|x|-ln|5x+1|+C

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integral of 1/((5x^2+x))

\int \frac{1}{( 5 x ^{2} + x )} d x

ln ∣ x ∣ - ln ∣ 5 x + 1 ∣ + C

Solution steps

\int \frac{1}{5 x ^{2} + x} d x

Take the partial fraction of

\frac{1}{5 x ^{2} + x} : \frac{1}{x} - \frac{5}{5 x + 1}

= \int \frac{1}{x} - \frac{5}{5 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{5}{5 x + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{5 x + 1} d x = ln ∣ 5 x + 1 ∣

= ln ∣ x ∣ - ln ∣ 5 x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ 5 x + 1 ∣ + C

\frac{d y}{d x} = 2 cos (5 x)

f (x) = 4 x + 5

x \to 2 + lim (\frac{∣ x - 2 ∣}{x - 2})

cos (x) - x^{2}

What is the integral of 1/((5x^2+x)) ?
The integral of 1/((5x^2+x)) is ln|x|-ln|5x+1|+C