What is the integral of (4x^2-2x+1)/(x^3+x) ?

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

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integral of (4x^2-2x+1)/(x^3+x)

\int \frac{4 x ^{2} - 2 x + 1}{x ^{3} + x} d x

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

Solution steps

\int \frac{4 x ^{2} - 2 x + 1}{x ^{3} + x} d x

Expand

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

\int \frac{4 x ^{2}}{x ^{3} + x} d x = 2 ln x^{2} + 1

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

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

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

Simplify

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

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

Add a constant to the solution

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

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