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

The integral of (x^2-6)/(6x) is (x^2)/(12)-ln|x|+C

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integral of (x^2-6)/(6x)

\int \frac{x ^{2} - 6}{6 x} d x

\frac{x ^{2}}{12} - ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 6}{6 x} 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{x ^{2} - 6}{x} d x

Expand

\frac{x ^{2} - 6}{x} : x - \frac{6}{x}

= \frac{1}{6} \cdot \int x - \frac{6}{x} 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 x d x - \int \frac{6}{x} d x)

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

= \frac{1}{6} (\frac{x ^{2}}{2} - 6 ln ∣ x ∣)

Simplify

\frac{1}{6} (\frac{x ^{2}}{2} - 6 ln ∣ x ∣) : \frac{x ^{2}}{12} - ln ∣ x ∣

= \frac{x ^{2}}{12} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{12} - ln ∣ x ∣ + C

\frac{\partial}{\partial s} (t + 3 t^{3} sin (s))

y = e^{3 x}, y = e^{6 x}, x = 1

\int x^{3} (7 - x^{2})^{\frac{1}{2}} d x

y^{^{'}} = e^{9 x} - 4 x

\int e^{- 2 x^{2}} \cdot (- 4) x^{3} d x

What is the integral of (x^2-6)/(6x) ?
The integral of (x^2-6)/(6x) is (x^2)/(12)-ln|x|+C