What is the integral of (x^4+9)/(x^4+9x^2) ?

The integral of (x^4+9)/(x^4+9x^2) is x-1/x-10/3 arctan(x/3)+C

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integral of (x^4+9)/(x^4+9x^2)

\int \frac{x ^{4} + 9}{x ^{4} + 9 x ^{2}} d x

x - \frac{1}{x} - \frac{10}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{x ^{4} + 9}{x ^{4} + 9 x ^{2}} d x

Take the partial fraction of

\frac{x ^{4} + 9}{x ^{4} + 9 x ^{2}} : 1 + \frac{1}{x ^{2}} - \frac{10}{x ^{2} + 9}

= \int 1 + \frac{1}{x ^{2}} - \frac{10}{x ^{2} + 9} 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 1 d x + \int \frac{1}{x ^{2}} d x - \int \frac{10}{x ^{2} + 9} d x

\int 1 d x = x

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

\int \frac{10}{x ^{2} + 9} d x = \frac{10}{3} arctan (\frac{x}{3})

= x - \frac{1}{x} - \frac{10}{3} arctan (\frac{x}{3})

Add a constant to the solution

= x - \frac{1}{x} - \frac{10}{3} arctan (\frac{x}{3}) + C

\frac{1}{( s + 6 ) ( s - 4 )}

\frac{d}{d x} (4^{(8 x^{4} + c o s (x))^{3}})

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

t^{2} - t \cdot y \cdot y^{^{'}} + y^{2} = 0

\frac{t ^{6}}{( t ^{3} + 1 ) ^{2}}

What is the integral of (x^4+9)/(x^4+9x^2) ?
The integral of (x^4+9)/(x^4+9x^2) is x-1/x-10/3 arctan(x/3)+C