What is the integral of (x^2-12)/(x^3+4x) ?

The integral of (x^2-12)/(x^3+4x) is 2ln|x^2+4|-3ln|x|+C

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integral of (x^2-12)/(x^3+4x)

\int \frac{x ^{2} - 12}{x ^{3} + 4 x} d x

2 ln x^{2} + 4 - 3 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 12}{x ^{3} + 4 x} d x

Expand

\frac{x ^{2} - 12}{x ^{3} + 4 x} : \frac{x ^{2}}{x ^{3} + 4 x} - \frac{12}{x ^{3} + 4 x}

Apply the Sum Rule:

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

= \int \frac{x ^{2}}{x ^{3} + 4 x} d x - \int \frac{12}{x ^{3} + 4 x} d x

\int \frac{x ^{2}}{x ^{3} + 4 x} d x = \frac{1}{2} ln x^{2} + 4

\int \frac{12}{x ^{3} + 4 x} d x = 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{1}{2} ln x^{2} + 4 - 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

\frac{1}{2} ln x^{2} + 4 - 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : 2 ln x^{2} + 4 - 3 ln ∣ x ∣

= 2 ln x^{2} + 4 - 3 ln ∣ x ∣

Add a constant to the solution

= 2 ln x^{2} + 4 - 3 ln ∣ x ∣ + C

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

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

\int \frac{1}{x 3 x} d x

\frac{d}{d x} (2 ln (x^{2} + y^{2}))

t y^{^{'}} = (1 - y^{2})^{\frac{1}{2}}

What is the integral of (x^2-12)/(x^3+4x) ?
The integral of (x^2-12)/(x^3+4x) is 2ln|x^2+4|-3ln|x|+C