What is the integral from 0 to 2 of x/((1+x^2)^2) ?

The integral from 0 to 2 of x/((1+x^2)^2) is 2/5

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integral from 0 to 2 of x/((1+x^2)^2)

\int_{0}^{2} \frac{x}{( 1 + x ^{2} ) ^{2}} d x

\frac{2}{5}

Decimal Notation

0.4

Solution steps

\int_{0}^{2} \frac{x}{( 1 + x ^{2} ) ^{2}} d x

Apply u-substitution

: \int_{1}^{5} \frac{1}{2 u ^{2}} d u

= \int_{1}^{5} \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{5}

= \frac{1}{2} [- \frac{1}{u}]_{1}^{5}

Compute the boundaries:

\frac{4}{5}

= \frac{1}{2} \cdot \frac{4}{5}

Simplify

= \frac{2}{5}

\frac{d}{d x} ((5 - 6 x)^{5})

2 x + \frac{3}{x}

x \to \infty lim (x e^{- 1})

\frac{\partial}{\partial x} (\frac{x}{1 + x ^{2}})

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

What is the integral from 0 to 2 of x/((1+x^2)^2) ?
The integral from 0 to 2 of x/((1+x^2)^2) is 2/5