What is the integral of (x^4+6x^2)/(x(x^2+4)^2) ?

The integral of (x^4+6x^2)/(x(x^2+4)^2) is 1/2 ln|x^2+4|-1/(x^2+4)+C

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integral of (x^4+6x^2)/(x(x^2+4)^2)

\int \frac{x ^{4} + 6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{1}{2} ln x^{2} + 4 - \frac{1}{x ^{2} + 4} + C

Solution steps

\int \frac{x ^{4} + 6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}} d x

Expand

\frac{x ^{4} + 6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}} : \frac{x ^{4}}{x ( x ^{2} + 4 ) ^{2}} + \frac{6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}}

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 ^{4}}{x ( x ^{2} + 4 ) ^{2}} d x + \int \frac{6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}} d x

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

\int \frac{6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}} d x = - \frac{3}{x ^{2} + 4}

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 - \frac{1}{x ^{2} + 4} + C

What is the integral of (x^4+6x^2)/(x(x^2+4)^2) ?
The integral of (x^4+6x^2)/(x(x^2+4)^2) is 1/2 ln|x^2+4|-1/(x^2+4)+C