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

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

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

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

\frac{3}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (ln x^{2} + 4 - arctan (\frac{x}{2})) + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{3}{5 ( x + 1 )} + \frac{2 x - 2}{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 \frac{3}{5 ( x + 1 )} d x + \int \frac{2 x - 2}{5 ( x ^{2} + 4 )} d x

\int \frac{3}{5 ( x + 1 )} d x = \frac{3}{5} ln ∣ x + 1 ∣

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

= \frac{3}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (ln x^{2} + 4 - arctan (\frac{x}{2}))

Add a constant to the solution

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

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