What is the integral of (x^3-2x^2+x+1)/(x^4+5x^2+4) ?

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

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integral of (x^3-2x^2+x+1)/(x^4+5x^2+4)

\int \frac{x ^{3} - 2 x ^{2} + x + 1}{x ^{4} + 5 x ^{2} + 4} d x

arctan (x) + \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{3} - 2 x ^{2} + x + 1}{x ^{4} + 5 x ^{2} + 4} d x

Take the partial fraction of

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

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

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

\int \frac{x - 3}{x ^{2} + 4} d x = \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2})

= arctan (x) + \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2})

Add a constant to the solution

= arctan (x) + \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) + C

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