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

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

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

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

diverges

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{6 ( u ^{2} - 1 )} d u

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}{u ^{2} - 1} d u

Factor

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

= \frac{1}{6} \cdot \int_{0}^{1} \frac{1}{- ( - u ^{2} + 1 )} d u

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}{- u ^{2} + 1} d u)

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

diverges

= diverges

\int e^{x} x - e^{x} d x

x \to 5 lim (x^{3})

\int \frac{1}{x ^{2} - 2 x + 5} d x

y^{^{''}} - y^{^{'}} - 2 y = 7 e^{- t}

y^{^{''}} + 16 y = 2 csc^{2} (4 t)

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