What is the integral of (x^4)/(25+4x^2) ?

The integral of (x^4)/(25+4x^2) is (x^3)/(12)-25/16 x+125/32 arctan(2/5 x)+C

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integral of (x^4)/(25+4x^2)

\int \frac{x ^{4}}{25 + 4 x ^{2}} d x

\frac{x ^{3}}{12} - \frac{25}{16} x + \frac{125}{32} arctan (\frac{2}{5} x) + C

Solution steps

\int \frac{x ^{4}}{25 + 4 x ^{2}} d x

Long division

\frac{x ^{4}}{25 + 4 x ^{2}} : \frac{x ^{2}}{4} - \frac{25}{16} + \frac{625}{16 ( 4 x ^{2} + 25 )}

= \int \frac{x ^{2}}{4} - \frac{25}{16} + \frac{625}{16 ( 4 x ^{2} + 25 )} 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 ^{2}}{4} d x - \int \frac{25}{16} d x + \int \frac{625}{16 ( 4 x ^{2} + 25 )} d x

\int \frac{x ^{2}}{4} d x = \frac{x ^{3}}{12}

\int \frac{25}{16} d x = \frac{25}{16} x

\int \frac{625}{16 ( 4 x ^{2} + 25 )} d x = \frac{125}{32} arctan (\frac{2}{5} x)

= \frac{x ^{3}}{12} - \frac{25}{16} x + \frac{125}{32} arctan (\frac{2}{5} x)

Add a constant to the solution

= \frac{x ^{3}}{12} - \frac{25}{16} x + \frac{125}{32} arctan (\frac{2}{5} x) + C

What is the integral of (x^4)/(25+4x^2) ?
The integral of (x^4)/(25+4x^2) is (x^3)/(12)-25/16 x+125/32 arctan(2/5 x)+C