What is the integral from 0 to 6 of 1/(u^2-36) ?

The integral from 0 to 6 of 1/(u^2-36) is diverges

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integral from 0 to 6 of 1/(u^2-36)

\int_{0}^{6} \frac{1}{u ^{2} - 36} d u

diverges

Solution steps

\int_{0}^{6} \frac{1}{u ^{2} - 36} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{6 ( 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}{6} \cdot \int_{0}^{1} \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{6} \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}{6} (- \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}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1})

Simplify

\frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{6} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{6} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

x \to 2 lim (\frac{∣ x - 2 ∣}{x - 2})

\int b cos (x) d x

\frac{d y}{d x} = \frac{y}{x} + 9 x + 4

\frac{\partial}{\partial x} (9 - x)

x \to 0 lim (\frac{x - ∣ x ∣}{2 x})

What is the integral from 0 to 6 of 1/(u^2-36) ?
The integral from 0 to 6 of 1/(u^2-36) is diverges