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

The integral from 0 to 1 of (2x)/((x^2+4)^2) is 1/20

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

\int_{0}^{1} \frac{2 x}{( x ^{2} + 4 ) ^{2}} d x

\frac{1}{20}

Decimal Notation

0.05

Solution steps

\int_{0}^{1} \frac{2 x}{( x ^{2} + 4 ) ^{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}^{1} \frac{x}{( x ^{2} + 4 ) ^{2}} d x

Apply u-substitution

: \int_{4}^{5} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{4}^{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_{4}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{4}^{5}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{5}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{4}^{5} : [- \frac{1}{u}]_{4}^{5}

= [- \frac{1}{u}]_{4}^{5}

Compute the boundaries:

\frac{1}{20}

= \frac{1}{20}

x, \frac{1}{4} x

y^{^{'''}} - y^{^{''}} = 0

\frac{\partial}{\partial x} (2 x^{2} y + x e^{y^{2}})

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

\int y + cos (x) d x

What is the integral from 0 to 1 of (2x)/((x^2+4)^2) ?
The integral from 0 to 1 of (2x)/((x^2+4)^2) is 1/20