What is the integral of (3x-7)/(x^3+x^2+4x+4) ?

The integral of (3x-7)/(x^3+x^2+4x+4) is -2ln|x+1|+ln|x^2+4|+1/2 arctan(x/2)+C

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integral of (3x-7)/(x^3+x^2+4x+4)

\int \frac{3 x - 7}{x ^{3} + x ^{2} + 4 x + 4} d x

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

Solution steps

\int \frac{3 x - 7}{x ^{3} + x ^{2} + 4 x + 4} d x

Take the partial fraction of

\frac{3 x - 7}{x ^{3} + x ^{2} + 4 x + 4} : - \frac{2}{x + 1} + \frac{2 x + 1}{x ^{2} + 4}

= \int - \frac{2}{x + 1} + \frac{2 x + 1}{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{2}{x + 1} d x + \int \frac{2 x + 1}{x ^{2} + 4} d x

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

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

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

Add a constant to the solution

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

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What is the integral of (3x-7)/(x^3+x^2+4x+4) ?
The integral of (3x-7)/(x^3+x^2+4x+4) is -2ln|x+1|+ln|x^2+4|+1/2 arctan(x/2)+C