What is the integral from 3 to 4 of x/((x^2+5)^3) ?

The integral from 3 to 4 of x/((x^2+5)^3) is 5/7056

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integral from 3 to 4 of x/((x^2+5)^3)

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

\frac{5}{7056}

Decimal

0.00070 \dots

Solution steps

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

Apply u-substitution

= \int_{14}^{21} \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_{14}^{21} \frac{1}{u ^{3}} d u

Apply the Power Rule

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{14}^{21}

Compute the boundaries:

\frac{5}{3528}

= \frac{1}{2} \cdot \frac{5}{3528}

Simplify

= \frac{5}{7056}

What is the integral from 3 to 4 of x/((x^2+5)^3) ?
The integral from 3 to 4 of x/((x^2+5)^3) is 5/7056