What is the integral of (2x^2+5x+16)/(x^4+8x^2+16) ?

The integral of (2x^2+5x+16)/(x^4+8x^2+16) is 3/2 arctan(x/2)+1/4 sin(2arctan(x/2))-5/(2(x^2+4))+C

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integral of (2x^2+5x+16)/(x^4+8x^2+16)

\int \frac{2 x ^{2} + 5 x + 16}{x ^{4} + 8 x ^{2} + 16} d x

\frac{3}{2} arctan (\frac{x}{2}) + \frac{1}{4} sin (2 arctan (\frac{x}{2})) - \frac{5}{2 ( x ^{2} + 4 )} + C

Solution steps

\int \frac{2 x ^{2} + 5 x + 16}{x ^{4} + 8 x ^{2} + 16} d x

Take the partial fraction of

\frac{2 x ^{2} + 5 x + 16}{x ^{4} + 8 x ^{2} + 16} : \frac{2}{x ^{2} + 4} + \frac{5 x + 8}{( x ^{2} + 4 ) ^{2}}

= \int \frac{2}{x ^{2} + 4} + \frac{5 x + 8}{( x ^{2} + 4 ) ^{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{2}{x ^{2} + 4} d x + \int \frac{5 x + 8}{( x ^{2} + 4 ) ^{2}} d x

\int \frac{2}{x ^{2} + 4} d x = arctan (\frac{x}{2})

\int \frac{5 x + 8}{( x ^{2} + 4 ) ^{2}} d x = - \frac{5}{2 ( x ^{2} + 4 )} + \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

= arctan (\frac{x}{2}) - \frac{5}{2 ( x ^{2} + 4 )} + \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

Simplify

arctan (\frac{x}{2}) - \frac{5}{2 ( x ^{2} + 4 )} + \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : \frac{3}{2} arctan (\frac{x}{2}) + \frac{1}{4} sin (2 arctan (\frac{x}{2})) - \frac{5}{2 ( x ^{2} + 4 )}

= \frac{3}{2} arctan (\frac{x}{2}) + \frac{1}{4} sin (2 arctan (\frac{x}{2})) - \frac{5}{2 ( x ^{2} + 4 )}

Add a constant to the solution

= \frac{3}{2} arctan (\frac{x}{2}) + \frac{1}{4} sin (2 arctan (\frac{x}{2})) - \frac{5}{2 ( x ^{2} + 4 )} + C

What is the integral of (2x^2+5x+16)/(x^4+8x^2+16) ?
The integral of (2x^2+5x+16)/(x^4+8x^2+16) is 3/2 arctan(x/2)+1/4 sin(2arctan(x/2))-5/(2(x^2+4))+C