What is the integral of ((x+3))/(2x^3-8x) ?

The integral of ((x+3))/(2x^3-8x) is 1/2 (-3/4 ln|x|+1/8 ln|x+2|+5/8 ln|x-2|)+C

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integral of ((x+3))/(2x^3-8x)

\int \frac{( x + 3 )}{2 x ^{3} - 8 x} d x

\frac{1}{2} (- \frac{3}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{5}{8} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{x + 3}{2 x ^{3} - 8 x} d x

Factor

\frac{x + 3}{2 x ^{3} - 8 x}

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

Take the constant out:

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

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

Take the partial fraction of

\frac{x + 3}{x ( x + 2 ) ( x - 2 )} : - \frac{3}{4 x} + \frac{1}{8 ( x + 2 )} + \frac{5}{8 ( x - 2 )}

= \frac{1}{2} \cdot \int - \frac{3}{4 x} + \frac{1}{8 ( x + 2 )} + \frac{5}{8 ( 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}{2} (- \int \frac{3}{4 x} d x + \int \frac{1}{8 ( x + 2 )} d x + \int \frac{5}{8 ( x - 2 )} d x)

\int \frac{3}{4 x} d x = \frac{3}{4} ln ∣ x ∣

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

\int \frac{5}{8 ( x - 2 )} d x = \frac{5}{8} ln ∣ x - 2 ∣

= \frac{1}{2} (- \frac{3}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{5}{8} ln ∣ x - 2 ∣)

Add a constant to the solution

= \frac{1}{2} (- \frac{3}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{5}{8} ln ∣ x - 2 ∣) + C

\frac{\partial}{\partial y} (x e^{z})

f (x) = 6 - 6 x

\frac{d}{d x} (r^{2} - x^{2})

n = 1 \sum \infty \frac{2}{2 ^{n} + 5}

\frac{d}{d x} (\frac{0.14 x}{0.15 + x})

What is the integral of ((x+3))/(2x^3-8x) ?
The integral of ((x+3))/(2x^3-8x) is 1/2 (-3/4 ln|x|+1/8 ln|x+2|+5/8 ln|x-2|)+C