What is the integral from 0 to 5 of 1/(u^2-25) ?

The integral from 0 to 5 of 1/(u^2-25) is diverges

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integral from 0 to 5 of 1/(u^2-25)

\int_{0}^{5} \frac{1}{u ^{2} - 25} d u

diverges

Solution steps

\int_{0}^{5} \frac{1}{u ^{2} - 25} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{5 ( 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}{5} \cdot \int_{0}^{1} \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{5} \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}{5} (- \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}{5} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1})

Simplify

\frac{1}{5} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

t an g e n t y = 4 cos (x), (\frac{π}{3}, 2)

d er i v a t i v e - x^{- 2} e^{\frac{1}{x}}

t an g e n t 5 + 4 x^{2} - 2 x^{3}

y^{^{''}} - 7 y^{^{'}} + 6 y = 0

\int \frac{y ^{3}}{y ^{2} + 1} d y

What is the integral from 0 to 5 of 1/(u^2-25) ?
The integral from 0 to 5 of 1/(u^2-25) is diverges