What is the integral of 1/(-x^2-a+b) ?

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

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integral of 1/(-x^2-a+b)

\int \frac{1}{- x ^{2} - a + b} d x

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

Solution steps

\int \frac{1}{- x ^{2} - a + b} d x

Apply Integral Substitution

= \int \frac{1}{( - u ^{2} + 1 ) - a + b} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{- a + b} \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}{- a + b} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{- a + b}

= \frac{1}{- a + b} \frac{ln \frac{x}{- a + b} + 1}{2} - \frac{ln \frac{x}{- a + b} - 1}{2}

Simplify

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

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

Add a constant to the solution

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

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

\frac{1}{3 s ^{2}} - \frac{5}{2 s}

y = 3 x, y = x

\frac{d}{d x} (ln (x^{4} 8 - 6 x))

g (x) = \frac{1 - 3 x}{5 + x}

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