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

The integral from 0 to 1 of (2x)/((x^2+3)^3) is 7/288

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

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

\frac{7}{288}

Decimal Notation

0.02430 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{7}{288}

= \frac{7}{288}

x \to \infty lim \frac{( e ^{x^{4} - x^{2}} )}{2}

f (x) = x 1 + x^{2}

y^{^{'}} + \frac{2}{t} y = 0

\int \frac{e ^{- 2 x}}{1 + e ^{- x}} d x

\frac{\partial}{\partial x} (3 x^{2} - 2 x y + x - 3 y)

What is the integral from 0 to 1 of (2x)/((x^2+3)^3) ?
The integral from 0 to 1 of (2x)/((x^2+3)^3) is 7/288