What is the integral of ((x^2-8))/(8x) ?

The integral of ((x^2-8))/(8x) is (x^2)/(16)-ln|x|+C

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integral of ((x^2-8))/(8x)

\int \frac{( x ^{2} - 8 )}{8 x} d x

\frac{x ^{2}}{16} - ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 8}{8 x} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{x ^{2} - 8}{x} d x

Expand

\frac{x ^{2} - 8}{x} : x - \frac{8}{x}

= \frac{1}{8} \cdot \int x - \frac{8}{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}{8} (\int x d x - \int \frac{8}{x} d x)

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

\int \frac{8}{x} d x = 8 ln ∣ x ∣

= \frac{1}{8} (\frac{x ^{2}}{2} - 8 ln ∣ x ∣)

Simplify

\frac{1}{8} (\frac{x ^{2}}{2} - 8 ln ∣ x ∣) : \frac{x ^{2}}{16} - ln ∣ x ∣

= \frac{x ^{2}}{16} - ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{16} - ln ∣ x ∣ + C

\int_{0}^{2} (x^{3} - 6 x^{2} + 8 x) d x

\int_{0}^{1} 6 cos (x^{2}) d x

\frac{\partial}{\partial x} (4 x^{3} - 4 y)

\frac{d y}{d x} + \frac{3 y}{x} = 6 x^{2}

sin (2 t)

What is the integral of ((x^2-8))/(8x) ?
The integral of ((x^2-8))/(8x) is (x^2)/(16)-ln|x|+C