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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{2 x ^{3} + x ^{2} + 4}{( x ^{2} + 4 ) ^{2}} : \frac{2 x + 1}{x ^{2} + 4} - \frac{8 x}{( x ^{2} + 4 ) ^{2}}

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x e^{y} - ln (x))

- \frac{1}{x ^{2}}

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

\frac{d}{d t} ((sin (t) + cos (t))^{2})

y = 5 x^{2} + \frac{6}{x}

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