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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x ^{2} + 3} d x = \frac{1}{3} arctan (\frac{x}{3})

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

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

Add a constant to the solution

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

\int e^{2 x} 1 + e^{x} d x

x \to 0 lim (\frac{x - sin ( x )}{x})

\frac{\partial}{\partial y} (y sin (y))

y = 8 x sin (x^{2})

\frac{\partial}{\partial y} (7 e^{7 e^{y x} + y} x)

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