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

The integral of 1/((x+1)^2(x^2+2)(x-3)) is

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

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

- \frac{11}{144} ln ∣ x + 1 ∣ + \frac{1}{12 ( x + 1 )} + \frac{1}{99} (\frac{7}{2} ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2})) + \frac{1}{176} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 1 ) ^{2} ( x ^{2} + 2 ) ( x - 3 )} : - \frac{11}{144 ( x + 1 )} - \frac{1}{12 ( x + 1 ) ^{2}} + \frac{7 x - 1}{99 ( x ^{2} + 2 )} + \frac{1}{176 ( x - 3 )}

= \int - \frac{11}{144 ( x + 1 )} - \frac{1}{12 ( x + 1 ) ^{2}} + \frac{7 x - 1}{99 ( x ^{2} + 2 )} + \frac{1}{176 ( 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{11}{144 ( x + 1 )} d x - \int \frac{1}{12 ( x + 1 ) ^{2}} d x + \int \frac{7 x - 1}{99 ( x ^{2} + 2 )} d x + \int \frac{1}{176 ( x - 3 )} d x

\int \frac{11}{144 ( x + 1 )} d x = \frac{11}{144} ln ∣ x + 1 ∣

\int \frac{1}{12 ( x + 1 ) ^{2}} d x = - \frac{1}{12 ( x + 1 )}

\int \frac{7 x - 1}{99 ( x ^{2} + 2 )} d x = \frac{1}{99} (\frac{7}{2} ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2}))

\int \frac{1}{176 ( x - 3 )} d x = \frac{1}{176} ln ∣ x - 3 ∣

= - \frac{11}{144} ln ∣ x + 1 ∣ - (- \frac{1}{12 ( x + 1 )}) + \frac{1}{99} (\frac{7}{2} ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2})) + \frac{1}{176} ln ∣ x - 3 ∣

Simplify

= - \frac{11}{144} ln ∣ x + 1 ∣ + \frac{1}{12 ( x + 1 )} + \frac{1}{99} (\frac{7}{2} ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2})) + \frac{1}{176} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{11}{144} ln ∣ x + 1 ∣ + \frac{1}{12 ( x + 1 )} + \frac{1}{99} (\frac{7}{2} ln x^{2} + 2 - \frac{1}{2} arctan (\frac{x}{2})) + \frac{1}{176} ln ∣ x - 3 ∣ + C

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