What is the integral from 1 to 2 of x/((x^2+4)^3) ?

The integral from 1 to 2 of x/((x^2+4)^3) is 39/6400

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integral from 1 to 2 of x/((x^2+4)^3)

\int_{1}^{2} \frac{x}{( x ^{2} + 4 ) ^{3}} d x

\frac{39}{6400}

Decimal

0.00609375

Solution steps

\int_{1}^{2} \frac{x}{( x ^{2} + 4 ) ^{3}} d x

Apply u-substitution

: \int_{5}^{8} \frac{1}{2 u ^{3}} d u

= \int_{5}^{8} \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

= \frac{1}{2} \cdot \int_{5}^{8} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{5}^{8}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{5}^{8}

Compute the boundaries:

\frac{39}{3200}

= \frac{1}{2} \cdot \frac{39}{3200}

Simplify

= \frac{39}{6400}

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What is the integral from 1 to 2 of x/((x^2+4)^3) ?
The integral from 1 to 2 of x/((x^2+4)^3) is 39/6400