What is the integral of (6x+5)/(6x^2) ?

The integral of (6x+5)/(6x^2) is ln|x|-5/(6x)+C

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integral of (6x+5)/(6x^2)

\int \frac{6 x + 5}{6 x ^{2}} d x

ln ∣ x ∣ - \frac{5}{6 x} + C

Solution steps

\int \frac{6 x + 5}{6 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{6 x + 5}{x ^{2}} d x

Expand

\frac{6 x + 5}{x ^{2}} : \frac{6}{x} + \frac{5}{x ^{2}}

= \frac{1}{6} \cdot \int \frac{6}{x} + \frac{5}{x ^{2}} 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

= \frac{1}{6} (\int \frac{6}{x} d x + \int \frac{5}{x ^{2}} d x)

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

= \frac{1}{6} (6 ln ∣ x ∣ - \frac{5}{x})

Simplify

\frac{1}{6} (6 ln ∣ x ∣ - \frac{5}{x}) : ln ∣ x ∣ - \frac{5}{6 x}

= ln ∣ x ∣ - \frac{5}{6 x}

Add a constant to the solution

= ln ∣ x ∣ - \frac{5}{6 x} + C

n = 1 \sum \infty - \frac{1}{n}

\frac{d}{d x} (x \cdot ln^{2} (x))

\int 4 cos (x) + 9 sin (x) d x

\int \frac{5}{( 3 x + 5 ) ^{4}} d x

f (x) = \frac{4 x + 1}{2 x ^{2} + 1}

What is the integral of (6x+5)/(6x^2) ?
The integral of (6x+5)/(6x^2) is ln|x|-5/(6x)+C