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

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

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

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

diverges

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{5 ( - 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}{5} \cdot \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}{5} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}

Simplify

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

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

Compute the boundaries:

diverges

= diverges

x \to - 1 - lim (\frac{9}{x ^{3} - 1})

\int \frac{3}{16 - x ^{2}} d x

f (x) = 4 t

x (x - 4)

n = 1 \sum \infty \frac{( - 1 ) ^{n} ( 2 ) ^{n - 1}}{( 2 ) ^{n + 1} ( n ) ^{\frac{1}{3}}}

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