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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

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}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d t} (tan (e^{6 t}) + e^{t a n (6 t)})

\frac{\partial}{\partial x} (cos (x y) + e^{yz} + ln (z x))

f^{9}

\frac{d y}{d x} = \frac{( 2 x + 1 )}{3 y}

\frac{d y}{d x} = 2 y - 4 x

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