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

The integral from 0 to 1 of (2x)/((x^2+2)^2) is 1/6

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

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{2}^{3} \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_{2}^{3} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{2}^{3}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{2}^{3}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{2}^{3} : [- \frac{1}{u}]_{2}^{3}

= [- \frac{1}{u}]_{2}^{3}

Compute the boundaries:

\frac{1}{6}

= \frac{1}{6}

\frac{\partial}{\partial y} (8 e^{x} ln (y))

\frac{\partial}{\partial x} (7 e^{5 x})

f (x) = e^{x^{2} - 4}

x \to 2 lim ((x + 3)^{2})

ln (x + e)

What is the integral from 0 to 1 of (2x)/((x^2+2)^2) ?
The integral from 0 to 1 of (2x)/((x^2+2)^2) is 1/6