What is the integral of 1/(10-x^2) ?

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

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integral of 1/(10-x^2)

\int \frac{1}{10 - x ^{2}} d x

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

Solution steps

\int \frac{1}{10 - x ^{2}} d x

Apply Integral Substitution

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

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

Add a constant to the solution

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

y^{^{'}} + 2 x y = x, y (0) = 3

\frac{d ^{2}}{d x ^{2}} ((cos (2 x)) (sin (6 x)))

\int \frac{cos ( x )}{sin ^{12} ( x )} d x

y = e

f (x) = \frac{4}{x - 1}

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