What is the integral of (8(x^2+4))/(x(x^2+8)) ?

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

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integral of (8(x^2+4))/(x(x^2+8))

\int \frac{8 ( x ^{2} + 4 )}{x ( x ^{2} + 8 )} d x

8 (\frac{1}{4} ln x^{2} + 8 + \frac{1}{2} ln ∣ x ∣) + C

Solution steps

\int \frac{8 ( x ^{2} + 4 )}{x ( x ^{2} + 8 )} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{2} + 4}{x ( x ^{2} + 8 )} d x

Expand

\frac{x ^{2} + 4}{x ( x ^{2} + 8 )} : \frac{x ^{2}}{x ( x ^{2} + 8 )} + \frac{4}{x ( x ^{2} + 8 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int \frac{x ^{2}}{x ( x ^{2} + 8 )} d x + \int \frac{4}{x ( x ^{2} + 8 )} d x)

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

\int \frac{4}{x ( x ^{2} + 8 )} d x = 4 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{16} ln x^{2} + 8)

= 8 (\frac{1}{2} ln x^{2} + 8 + 4 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{16} ln x^{2} + 8))

Expand

\frac{1}{2} ln x^{2} + 8 + 4 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{16} ln x^{2} + 8) : \frac{1}{4} ln x^{2} + 8 + \frac{1}{2} ln ∣ x ∣

= 8 (\frac{1}{4} ln x^{2} + 8 + \frac{1}{2} ln ∣ x ∣)

Add a constant to the solution

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

\frac{d}{d x} (5 x^{3} + 3 x^{2} - 4 x + 1)

x \to 7 + lim (\frac{1}{x - 7})

\int e^{2 x + 4} d x

\frac{\partial}{\partial x} (- x (2 x - y^{3}))

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