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

The integral from 0 to 2 of (2x)/((1+x^2)^2) is 4/5

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

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

\frac{4}{5}

Decimal Notation

0.8

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{4}{5}

= \frac{4}{5}

\int 3 x^{3} 4 x^{4} d x

x \to 1 lim (2 x^{3} - 4 x^{2} + x + 8)

9 x + 4

y^{^{''}} + 2 y^{^{'}} + 5 y = 5 sin (2 t)

y = (e^{x} + 4)^{3}

What is the integral from 0 to 2 of (2x)/((1+x^2)^2) ?
The integral from 0 to 2 of (2x)/((1+x^2)^2) is 4/5