What is the integral of (7x^5-8x^2+1)/((x+1)(x^2+4)) ?

The integral of (7x^5-8x^2+1)/((x+1)(x^2+4)) is

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integral of (7x^5-8x^2+1)/((x+1)(x^2+4))

\int \frac{7 x ^{5} - 8 x ^{2} + 1}{( x + 1 ) ( x ^{2} + 4 )} d x

\frac{7 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 21 x - \frac{14}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (\frac{79}{2} ln x^{2} + 4 + \frac{481}{2} arctan (\frac{x}{2})) + C

Solution steps

\int \frac{7 x ^{5} - 8 x ^{2} + 1}{( x + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{7 x ^{5} - 8 x ^{2} + 1}{( x + 1 ) ( x ^{2} + 4 )} : 7 x^{2} - 7 x - 21 - \frac{14}{5 ( x + 1 )} + \frac{79 x + 481}{5 ( x ^{2} + 4 )}

= \int 7 x^{2} - 7 x - 21 - \frac{14}{5 ( x + 1 )} + \frac{79 x + 481}{5 ( x ^{2} + 4 )} 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 7 x^{2} d x - \int 7 x d x - \int 21 d x - \int \frac{14}{5 ( x + 1 )} d x + \int \frac{79 x + 481}{5 ( x ^{2} + 4 )} d x

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 21 d x = 21 x

\int \frac{14}{5 ( x + 1 )} d x = \frac{14}{5} ln ∣ x + 1 ∣

\int \frac{79 x + 481}{5 ( x ^{2} + 4 )} d x = \frac{1}{5} (\frac{79}{2} ln x^{2} + 4 + \frac{481}{2} arctan (\frac{x}{2}))

= \frac{7 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 21 x - \frac{14}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (\frac{79}{2} ln x^{2} + 4 + \frac{481}{2} arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{7 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 21 x - \frac{14}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (\frac{79}{2} ln x^{2} + 4 + \frac{481}{2} arctan (\frac{x}{2})) + C

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What is the integral of (7x^5-8x^2+1)/((x+1)(x^2+4)) ?
The integral of (7x^5-8x^2+1)/((x+1)(x^2+4)) is