What is the integral of (x^4(1-x)^4)/(1+x^2) ?

The integral of (x^4(1-x)^4)/(1+x^2) is (x^7)/7-(2x^6)/3+x^5-(4x^3)/3+4x-4arctan(x)+C

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integral of (x^4(1-x)^4)/(1+x^2)

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

\frac{x ^{7}}{7} - \frac{2 x ^{6}}{3} + x^{5} - \frac{4 x ^{3}}{3} + 4 x - 4 arctan (x) + C

Solution steps

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

Long division

\frac{x ^{4} ( 1 - x ) ^{4}}{1 + x ^{2}} : x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{x ^{2} + 1}

= \int x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{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^{6} d x - \int 4 x^{5} d x + \int 5 x^{4} d x - \int 4 x^{2} d x + \int 4 d x - \int \frac{4}{x ^{2} + 1} d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int 5 x^{4} d x = x^{5}

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

\int 4 d x = 4 x

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

= \frac{x ^{7}}{7} - \frac{2 x ^{6}}{3} + x^{5} - \frac{4 x ^{3}}{3} + 4 x - 4 arctan (x)

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{2 x ^{6}}{3} + x^{5} - \frac{4 x ^{3}}{3} + 4 x - 4 arctan (x) + C

\int (2 x + 1)^{20} d x

\int (9 cos (x)) d x

\int (x e^{- x})^{2} d x

\int e^{20 t} d t

What is the integral of (x^4(1-x)^4)/(1+x^2) ?
The integral of (x^4(1-x)^4)/(1+x^2) is (x^7)/7-(2x^6)/3+x^5-(4x^3)/3+4x-4arctan(x)+C