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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 2 \cdot \int - \frac{1}{x ^{2} + 1} - \frac{1}{( x ^{2} + 1 ) ^{2}} + \frac{1}{x ^{2}} 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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{3} (5.4)^{x})

\frac{d}{d x} (x e^{2 y})

n = 0 \sum \infty (- 3 x + 1)^{n}

\int_{0}^{\frac{π}{9}} cos (3 x) d x

\frac{d}{d x} (\frac{1}{2} cos (2 x))

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