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

The integral from 0 to 1 of (2x)/((x^2+2)^3) is 5/72

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

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

\frac{5}{72}

Decimal Notation

0.06944 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{5}{72}

= \frac{5}{72}

\frac{\partial}{\partial y} (tan (4 + 2 x^{2} y^{3} z^{2}))

\int \frac{( x ^{4} - 3 x ^{2} + 5 )}{x ^{4}} d x

\int (ln (x))^{3} \frac{1}{x} d x

10 x^{1.5} - 6 x^{0.5}

\frac{( 2 )}{( s ^{2} + 40 s + 400 )}

What is the integral from 0 to 1 of (2x)/((x^2+2)^3) ?
The integral from 0 to 1 of (2x)/((x^2+2)^3) is 5/72