What is the integral of (3x^2+3x+1)/(x^3+x) ?

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

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integral of (3x^2+3x+1)/(x^3+x)

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

ln x^{2} + 1 + 3 arctan (x) + ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{3 x ^{2} + 3 x + 1}{x ^{3} + x} : \frac{3 x ^{2}}{x ^{3} + x} + \frac{3 x}{x ^{3} + x} + \frac{1}{x ^{3} + 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 x ^{2}}{x ^{3} + x} d x + \int \frac{3 x}{x ^{3} + x} d x + \int \frac{1}{x ^{3} + x} d x

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

\int \frac{3 x}{x ^{3} + x} d x = 3 arctan (x)

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

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

Simplify

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

= ln x^{2} + 1 + 3 arctan (x) + ln ∣ x ∣

Add a constant to the solution

= ln x^{2} + 1 + 3 arctan (x) + ln ∣ x ∣ + C

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