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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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