What is the integral from 4 to 8 of z/(sqrt(z^2-15)) ?

The integral from 4 to 8 of z/(sqrt(z^2-15)) is 6

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integral from 4 to 8 of z/(sqrt(z^2-15))

\int_{4}^{8} \frac{z}{z ^{2} - 15} d z

6

Solution steps

\int_{4}^{8} \frac{z}{z ^{2} - 15} d z

Apply u-substitution

: \int_{1}^{49} \frac{1}{2 u} d u

= \int_{1}^{49} \frac{1}{2 u} 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_{1}^{49} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{49}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{49}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{1}^{49}

Compute the boundaries:

12

= \frac{1}{2} \cdot 12

Simplify

= 6

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What is the integral from 4 to 8 of z/(sqrt(z^2-15)) ?
The integral from 4 to 8 of z/(sqrt(z^2-15)) is 6