What is the integral from 0 to 2 of (2x)/((x^2+5)^3) ?

The integral from 0 to 2 of (2x)/((x^2+5)^3) is 28/2025

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integral from 0 to 2 of (2x)/((x^2+5)^3)

\int_{0}^{2} \frac{2 x}{( x ^{2} + 5 ) ^{3}} d x

\frac{28}{2025}

Decimal Notation

0.01382 \dots

Solution steps

\int_{0}^{2} \frac{2 x}{( x ^{2} + 5 ) ^{3}} 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}{( x ^{2} + 5 ) ^{3}} d x

Apply u-substitution

: \int_{5}^{9} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{5}^{9} \frac{1}{2 u ^{3}} 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_{5}^{9} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{5}^{9}

= 2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{5}^{9}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{5}^{9} : [- \frac{1}{2 u ^{2}}]_{5}^{9}

= [- \frac{1}{2 u ^{2}}]_{5}^{9}

Compute the boundaries:

\frac{28}{2025}

= \frac{28}{2025}

x \to 0 lim (\frac{- 2 e ^{- x}}{2 x - 2})

\int \frac{e ^{4 x}}{- 4 x} d x

x \to + 0 lim (\frac{( 3 e ^{x} - 3 )}{x})

\int_{0}^{1} 0.5 e^{- 2 x} d x

\int 2 xx d x

What is the integral from 0 to 2 of (2x)/((x^2+5)^3) ?
The integral from 0 to 2 of (2x)/((x^2+5)^3) is 28/2025