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

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

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

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

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

Solution steps

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

Long division

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

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

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\int x (x^{2} - 8)^{3} d x

\int sin^{5} (x) cot (x) d x

d er i v a t i v e \frac{ln ( t )}{t}

\int \frac{x ^{2} - 256}{x} d x

a re a x = - 3, x = 4, y = e^{2 x}, y = 2 e^{- x}

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