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

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

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

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

\frac{1}{4} ln x^{4} + 1 + \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 2 arctan (x^{2}) + C

Solution steps

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

Expand

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

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

Expand

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

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

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

\int \frac{4 x ^{2}}{x ^{4} + 1} d x = - \frac{1}{2} ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)

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

= \frac{1}{4} ln x^{4} + 1 - (- \frac{1}{2} ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) + 2 arctan (x^{2})

Simplify

= \frac{1}{4} ln x^{4} + 1 + \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 2 arctan (x^{2})

Add a constant to the solution

= \frac{1}{4} ln x^{4} + 1 + \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 2 arctan (x^{2}) + C

\int \frac{cos ^{2} ( x )}{cos ( x )} d x

f^{^{'}} (x) = \frac{x ^{2}}{x + 1}

s l o p e (- 4, 1), (- 7, - 5)

x y + \frac{d y}{d x} = 120 x

\int ln (6 - 3 x) d x

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