What is the integral of (x^6)/((1+x^2)^2) ?

The integral of (x^6)/((1+x^2)^2) is (x^3)/3-2x-(3x)/(2(x^2+1))+5/2 arctan(x)+x/(x^2+1)+C

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integral of (x^6)/((1+x^2)^2)

\int \frac{x ^{6}}{( 1 + x ^{2} ) ^{2}} d x

\frac{x ^{3}}{3} - 2 x - \frac{3 x}{2 ( x ^{2} + 1 )} + \frac{5}{2} arctan (x) + \frac{x}{x ^{2} + 1} + C

Solution steps

\int \frac{x ^{6}}{( 1 + x ^{2} ) ^{2}} d x

Long division

\frac{x ^{6}}{( 1 + x ^{2} ) ^{2}} : x^{2} - 2 + \frac{3 x ^{2} + 2}{x ^{4} + 2 x ^{2} + 1}

= \int x^{2} - 2 + \frac{3 x ^{2} + 2}{x ^{4} + 2 x ^{2} + 1} 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 x^{2} d x - \int 2 d x + \int \frac{3 x ^{2} + 2}{x ^{4} + 2 x ^{2} + 1} d x

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

\int 2 d x = 2 x

\int \frac{3 x ^{2} + 2}{x ^{4} + 2 x ^{2} + 1} d x = - \frac{3 x}{2 ( x ^{2} + 1 )} + \frac{5}{2} arctan (x) + \frac{x}{x ^{2} + 1}

= \frac{x ^{3}}{3} - 2 x - \frac{3 x}{2 ( x ^{2} + 1 )} + \frac{5}{2} arctan (x) + \frac{x}{x ^{2} + 1}

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x - \frac{3 x}{2 ( x ^{2} + 1 )} + \frac{5}{2} arctan (x) + \frac{x}{x ^{2} + 1} + C

(1 - y^{2}) d x - x y d y = 0

\int (x^{2} + 1)^{2} 2 x d x

y^{^{'}} = \frac{t}{y e ^{- y^{2}} + y e ^{- y^{2}} t ^{2}}

\frac{d}{d x} (- \frac{14}{x ^{4}})

\int_{- \frac{π}{2}}^{\frac{π}{2}} tan (x) d x

What is the integral of (x^6)/((1+x^2)^2) ?
The integral of (x^6)/((1+x^2)^2) is (x^3)/3-2x-(3x)/(2(x^2+1))+5/2 arctan(x)+x/(x^2+1)+C