What is the integral of 1/(100-x^2) ?

The integral of 1/(100-x^2) is 1/20 (ln| x/(10)+1|-ln| x/(10)-1|)+C

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integral of 1/(100-x^2)

\int \frac{1}{100 - x ^{2}} d x

\frac{1}{20} (ln \frac{x}{10} + 1 - ln \frac{x}{10} - 1) + C

Solution steps

\int \frac{1}{100 - x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{10 ( - 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}{10} \cdot \int \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}{10} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{10}

= \frac{1}{10} (\frac{ln \frac{x}{10} + 1}{2} - \frac{ln \frac{x}{10} - 1}{2})

Simplify

\frac{1}{10} (\frac{ln \frac{x}{10} + 1}{2} - \frac{ln \frac{x}{10} - 1}{2}) : \frac{1}{20} (ln \frac{x}{10} + 1 - ln \frac{x}{10} - 1)

= \frac{1}{20} (ln \frac{x}{10} + 1 - ln \frac{x}{10} - 1)

Add a constant to the solution

= \frac{1}{20} (ln \frac{x}{10} + 1 - ln \frac{x}{10} - 1) + C

x \to 3 lim (2 (x + 3))

\int x \cdot sin (3 x^{2}) d x

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

\int \frac{( cos ^{3} ( x ) )}{sin ( x )} d x

\int 10 y d y

What is the integral of 1/(100-x^2) ?
The integral of 1/(100-x^2) is 1/20 (ln| x/(10)+1|-ln| x/(10)-1|)+C