What is the integral from 0 to 7 of 1/(u^2-49) ?

The integral from 0 to 7 of 1/(u^2-49) is diverges

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integral from 0 to 7 of 1/(u^2-49)

\int_{0}^{7} \frac{1}{u ^{2} - 49} d u

diverges

Solution steps

\int_{0}^{7} \frac{1}{u ^{2} - 49} d u

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{7 ( 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}{7} \cdot \int_{0}^{1} \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{7} \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}{7} (- \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}{7} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1})

Simplify

\frac{1}{7} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{7} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{7} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (\frac{2000}{x ^{2}})

\frac{d x}{d t} = x^{2} t^{2}

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

\int e^{3 c o s (2 x)} sin (2 x) d x

t^{2} sin (3 t)

What is the integral from 0 to 7 of 1/(u^2-49) ?
The integral from 0 to 7 of 1/(u^2-49) is diverges