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

The integral of (x-7)/((x+1)(x^2+2x+5)) is -2ln|x+1|+ln|x^2+2x+5|+1/2 arctan(1/2 (x+1))+C

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

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

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

Solution steps

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

Take the partial fraction of

\frac{x - 7}{( x + 1 ) ( x ^{2} + 2 x + 5 )} : - \frac{2}{x + 1} + \frac{2 x + 3}{x ^{2} + 2 x + 5}

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

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

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

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

Add a constant to the solution

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

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What is the integral of (x-7)/((x+1)(x^2+2x+5)) ?
The integral of (x-7)/((x+1)(x^2+2x+5)) is -2ln|x+1|+ln|x^2+2x+5|+1/2 arctan(1/2 (x+1))+C