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

\int \frac{1}{( 2 x + 3 )} d x

\int x^{3} 1 - x^{2} d x

\frac{d}{d x} ((x + x)^{2})

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{6}

\int \frac{1}{- x ^{2} - a + b} d x

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