What is the integral of (6x^5)/((5-x^2)^2) ?

The integral of (6x^5)/((5-x^2)^2) is

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integral of (6x^5)/((5-x^2)^2)

\int \frac{6 x ^{5}}{( 5 - x ^{2} ) ^{2}} d x

6 (\frac{x ^{2}}{2} + 10 (\frac{x ^{4}}{4 ( x ^{4} - 10 x ^{2} + 25 )} + \frac{1}{2} ln x + 5 + \frac{7 5}{16 ( x + 5 )} - \frac{5}{16 ( x + 5 ) ^{2}} + \frac{1}{2} ln x - 5 - \frac{7 5}{16 ( x - 5 )} - \frac{5}{16 ( x - 5 ) ^{2}}) - \frac{5 5}{2 3} arctan (\frac{x ^{2}}{15})) + C

Solution steps

\int \frac{6 x ^{5}}{( 5 - x ^{2} ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{x ^{5}}{( 5 - x ^{2} ) ^{2}} d x

Long division

\frac{x ^{5}}{( 5 - x ^{2} ) ^{2}} : x + \frac{10 x ^{3} - 25 x}{x ^{4} - 10 x ^{2} + 25}

= 6 \cdot \int x + \frac{10 x ^{3} - 25 x}{x ^{4} - 10 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

= 6 (\int x d x + \int \frac{10 x ^{3} - 25 x}{x ^{4} - 10 x ^{2} + 25} d x)

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

\int \frac{10 x ^{3} - 25 x}{x ^{4} - 10 x ^{2} + 25} d x = 10 (\frac{x ^{4}}{4 ( x ^{4} - 10 x ^{2} + 25 )} + \frac{1}{2} ln x + 5 + \frac{7 5}{16 ( x + 5 )} - \frac{5}{16 ( x + 5 ) ^{2}} + \frac{1}{2} ln x - 5 - \frac{7 5}{16 ( x - 5 )} - \frac{5}{16 ( x - 5 ) ^{2}}) - \frac{5 5}{2 3} arctan (\frac{x ^{2}}{15})

= 6 (\frac{x ^{2}}{2} + 10 (\frac{x ^{4}}{4 ( x ^{4} - 10 x ^{2} + 25 )} + \frac{1}{2} ln x + 5 + \frac{7 5}{16 ( x + 5 )} - \frac{5}{16 ( x + 5 ) ^{2}} + \frac{1}{2} ln x - 5 - \frac{7 5}{16 ( x - 5 )} - \frac{5}{16 ( x - 5 ) ^{2}}) - \frac{5 5}{2 3} arctan (\frac{x ^{2}}{15}))

Add a constant to the solution

= 6 (\frac{x ^{2}}{2} + 10 (\frac{x ^{4}}{4 ( x ^{4} - 10 x ^{2} + 25 )} + \frac{1}{2} ln x + 5 + \frac{7 5}{16 ( x + 5 )} - \frac{5}{16 ( x + 5 ) ^{2}} + \frac{1}{2} ln x - 5 - \frac{7 5}{16 ( x - 5 )} - \frac{5}{16 ( x - 5 ) ^{2}}) - \frac{5 5}{2 3} arctan (\frac{x ^{2}}{15})) + C

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What is the integral of (6x^5)/((5-x^2)^2) ?
The integral of (6x^5)/((5-x^2)^2) is