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

The integral from 1 to 2 of (2x)/((x^2+2)^3) is 1/24

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

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

\frac{1}{24}

Decimal

0.04166 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{6} \frac{1}{2 u ^{3}} d u

= 2 \cdot \int_{3}^{6} \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}^{6} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{3}^{6}

= 2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{3}^{6}

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{3}^{6} : [- \frac{1}{2 u ^{2}}]_{3}^{6}

= [- \frac{1}{2 u ^{2}}]_{3}^{6}

Compute the boundaries:

\frac{1}{24}

= \frac{1}{24}

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What is the integral from 1 to 2 of (2x)/((x^2+2)^3) ?
The integral from 1 to 2 of (2x)/((x^2+2)^3) is 1/24