What is the integral of (x^4)/((x+1)(x+2)(x+3)) ?

The integral of (x^4)/((x+1)(x+2)(x+3)) is (x^2)/2-6x+1/2 ln|x+1|-16ln|x+2|+81/2 ln|x+3|+C

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integral of (x^4)/((x+1)(x+2)(x+3))

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

\frac{x ^{2}}{2} - 6 x + \frac{1}{2} ln ∣ x + 1 ∣ - 16 ln ∣ x + 2 ∣ + \frac{81}{2} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4}}{( x + 1 ) ( x + 2 ) ( x + 3 )} : x - 6 + \frac{1}{2 ( x + 1 )} - \frac{16}{x + 2} + \frac{81}{2 ( x + 3 )}

= \int x - 6 + \frac{1}{2 ( x + 1 )} - \frac{16}{x + 2} + \frac{81}{2 ( 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 x d x - \int 6 d x + \int \frac{1}{2 ( x + 1 )} d x - \int \frac{16}{x + 2} d x + \int \frac{81}{2 ( x + 3 )} d x

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

\int 6 d x = 6 x

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

\int \frac{16}{x + 2} d x = 16 ln ∣ x + 2 ∣

\int \frac{81}{2 ( x + 3 )} d x = \frac{81}{2} ln ∣ x + 3 ∣

= \frac{x ^{2}}{2} - 6 x + \frac{1}{2} ln ∣ x + 1 ∣ - 16 ln ∣ x + 2 ∣ + \frac{81}{2} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 6 x + \frac{1}{2} ln ∣ x + 1 ∣ - 16 ln ∣ x + 2 ∣ + \frac{81}{2} ln ∣ x + 3 ∣ + C

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What is the integral of (x^4)/((x+1)(x+2)(x+3)) ?
The integral of (x^4)/((x+1)(x+2)(x+3)) is (x^2)/2-6x+1/2 ln|x+1|-16ln|x+2|+81/2 ln|x+3|+C