What is the integral of x/(x^3+3x^2+7x+5) ?

The integral of x/(x^3+3x^2+7x+5) is -1/4 ln|x+1|+1/4 (1/2 ln| 1/4 (x+1)^2+1|+2arctan(1/2 (x+1)))+C

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integral of x/(x^3+3x^2+7x+5)

\int \frac{x}{x ^{3} + 3 x ^{2} + 7 x + 5} d x

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

Solution steps

\int \frac{x}{x ^{3} + 3 x ^{2} + 7 x + 5} d x

Take the partial fraction of

\frac{x}{x ^{3} + 3 x ^{2} + 7 x + 5} : - \frac{1}{4 ( x + 1 )} + \frac{x + 5}{4 ( x ^{2} + 2 x + 5 )}

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

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What is the integral of x/(x^3+3x^2+7x+5) ?
The integral of x/(x^3+3x^2+7x+5) is -1/4 ln|x+1|+1/4 (1/2 ln| 1/4 (x+1)^2+1|+2arctan(1/2 (x+1)))+C