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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{x - 1}, 10

y^{^{''}} + π^{2} y = 2

2 x^{3} - 2 x^{2} + 12

x^{\frac{3}{8}}

x \to 0 lim (- 2 - x)

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