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

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

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integral of 1/(x^2-y)

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{y ( - 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}{y} \cdot \int \frac{1}{- 1 + u ^{2}} d u

Factor

- 1 + u^{2} : - (1 - u^{2})

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

Substitute back

u = \frac{x}{y}

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

Simplify

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

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

Add a constant to the solution

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

f (x) = tan (x)

- \frac{1}{2} cos^{2} (x)

y = x^{2} - 3 x - 4, (2, - 6)

\int_{0}^{2} 5^{x} d x

\int arccot (x) d x

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