What is the integral of 1/(1-25x^2) ?

The integral of 1/(1-25x^2) is 1/10 (ln|5x+1|-ln|5x-1|)+C

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integral of 1/(1-25x^2)

\int \frac{1}{1 - 25 x ^{2}} d x

\frac{1}{10} (ln ∣ 5 x + 1 ∣ - ln ∣ 5 x - 1 ∣) + C

Solution steps

\int \frac{1}{1 - 25 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 = 5 x

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

Simplify

\frac{1}{5} (\frac{ln ∣ 5 x + 1 ∣}{2} - \frac{ln ∣ 5 x - 1 ∣}{2}) : \frac{1}{10} (ln ∣ 5 x + 1 ∣ - ln ∣ 5 x - 1 ∣)

= \frac{1}{10} (ln ∣ 5 x + 1 ∣ - ln ∣ 5 x - 1 ∣)

Add a constant to the solution

= \frac{1}{10} (ln ∣ 5 x + 1 ∣ - ln ∣ 5 x - 1 ∣) + C

\frac{d}{d x} (\frac{- 5}{4 x ^{3}})

in v erse l a pl a ce \frac{5 s + 5}{s ( s ^{2} + 6 s + 10 )}

x \to - 1 lim (- \frac{2}{( x + 1 ) ^{2}})

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

in v erse l a pl a ce \frac{2}{s + 1} + \frac{24 s}{s ^{2} + 9}

What is the integral of 1/(1-25x^2) ?
The integral of 1/(1-25x^2) is 1/10 (ln|5x+1|-ln|5x-1|)+C