What is the integral from 0 to 1 of (2x)/((x^2+3)^2) ?

The integral from 0 to 1 of (2x)/((x^2+3)^2) is 1/12

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integral from 0 to 1 of (2x)/((x^2+3)^2)

\int_{0}^{1} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x

\frac{1}{12}

Decimal Notation

0.08333 \dots

Solution steps

\int_{0}^{1} \frac{2 x}{( x ^{2} + 3 ) ^{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} + 3 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{4} \frac{1}{2 u ^{2}} d u

= 2 \cdot \int_{3}^{4} \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_{3}^{4} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{3}^{4}

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{4}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{4} : [- \frac{1}{u}]_{3}^{4}

= [- \frac{1}{u}]_{3}^{4}

Compute the boundaries:

\frac{1}{12}

= \frac{1}{12}

f (x) = \frac{- 9 x ^{2}}{2} + 7 x + 2

\frac{\partial}{\partial y} (arctan (x^{4} y^{3}))

5 (t - 1)

\frac{d}{d x} (\frac{A}{B + C e ^{x}})

\int e^{a x} d x

What is the integral from 0 to 1 of (2x)/((x^2+3)^2) ?
The integral from 0 to 1 of (2x)/((x^2+3)^2) is 1/12