What is the integral from 0 to 4 of 1/(u^2-16) ?

The integral from 0 to 4 of 1/(u^2-16) is diverges

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integral from 0 to 4 of 1/(u^2-16)

\int_{0}^{4} \frac{1}{u ^{2} - 16} d u

diverges

Solution steps

\int_{0}^{4} \frac{1}{u ^{2} - 16} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{4 ( v ^{2} - 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} (- \int_{0}^{1} \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{4} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1})

Simplify

\frac{1}{4} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{4} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{4} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

n = 1 \sum \infty \frac{1}{n + n ^{2}}

\frac{\partial}{\partial x} (6 x^{8} y^{6} + 2 x^{7} y^{4})

f (x) = - tan (x) - 5 cot (x)

\frac{d}{d x} (x (x - 1)^{3})

\int - x^{2} (x + 1)^{3} (x - 1) d x

What is the integral from 0 to 4 of 1/(u^2-16) ?
The integral from 0 to 4 of 1/(u^2-16) is diverges