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

The integral of x/((x^2-4)(x+3)) is 1/2 ln|x+2|+1/10 ln|x-2|-3/5 ln|x+3|+C

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

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

\frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{10} ln ∣ x - 2 ∣ - \frac{3}{5} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x ^{2} - 4 ) ( x + 3 )} : \frac{1}{2 ( x + 2 )} + \frac{1}{10 ( x - 2 )} - \frac{3}{5 ( x + 3 )}

= \int \frac{1}{2 ( x + 2 )} + \frac{1}{10 ( x - 2 )} - \frac{3}{5 ( x + 3 )} 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

= \int \frac{1}{2 ( x + 2 )} d x + \int \frac{1}{10 ( x - 2 )} d x - \int \frac{3}{5 ( x + 3 )} d x

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

\int \frac{1}{10 ( x - 2 )} d x = \frac{1}{10} ln ∣ x - 2 ∣

\int \frac{3}{5 ( x + 3 )} d x = \frac{3}{5} ln ∣ x + 3 ∣

= \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{10} ln ∣ x - 2 ∣ - \frac{3}{5} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{10} ln ∣ x - 2 ∣ - \frac{3}{5} ln ∣ x + 3 ∣ + C

y = 1 - e^{x}

\int 1 5^{x} d x

\frac{d}{d x} ((1 + 2 x)^{2})

\frac{1}{x ^{2}}

\int \frac{x + 1}{x ^{2} + 1} d x

What is the integral of x/((x^2-4)(x+3)) ?
The integral of x/((x^2-4)(x+3)) is 1/2 ln|x+2|+1/10 ln|x-2|-3/5 ln|x+3|+C