What is the integral of 1/(5-x^2) ?

The integral of 1/(5-x^2) is 1/(2sqrt(5))(ln| x/(sqrt(5))+1|-ln| x/(sqrt(5))-1|)+C

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integral of 1/(5-x^2)

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

\frac{1}{2 5} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{5 ( 1 - u ^{2} )} 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 \frac{1}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} 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})

Substitute back

u = \frac{x}{5}

= \frac{1}{5} \frac{ln \frac{x}{5} + 1}{2} - \frac{ln \frac{x}{5} - 1}{2}

Simplify

\frac{1}{5} \frac{ln \frac{x}{5} + 1}{2} - \frac{ln \frac{x}{5} - 1}{2} : \frac{1}{2 5} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1)

= \frac{1}{2 5} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1)

Add a constant to the solution

= \frac{1}{2 5} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1) + C

\frac{d}{d x} (20 x^{3} - 6 x^{2} + x)

\frac{1}{s ^{2} - 4 s + 4}

\frac{d}{d x} (\frac{2}{x y})

\int_{0}^{\frac{π}{4}} 2 sec (x) d x

f^{^{'}} (x) = x^{2} e^{x^{2}}

What is the integral of 1/(5-x^2) ?
The integral of 1/(5-x^2) is 1/(2sqrt(5))(ln| x/(sqrt(5))+1|-ln| x/(sqrt(5))-1|)+C