What is the integral of 1/(u^2-3) ?

The integral of 1/(u^2-3) is -1/(2sqrt(3))(ln| u/(sqrt(3))+1|-ln| u/(sqrt(3))-1|)+C

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integral of 1/(u^2-3)

\int \frac{1}{u ^{2} - 3} d u

- \frac{1}{2 3} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1) + C

Solution steps

\int \frac{1}{u ^{2} - 3} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} - 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{3} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} (- \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}{3} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

v = \frac{u}{3}

= \frac{1}{3} - \frac{ln \frac{u}{3} + 1}{2} - \frac{ln \frac{u}{3} - 1}{2}

Simplify

\frac{1}{3} - \frac{ln \frac{u}{3} + 1}{2} - \frac{ln \frac{u}{3} - 1}{2} : - \frac{1}{2 3} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

= - \frac{1}{2 3} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

Add a constant to the solution

= - \frac{1}{2 3} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1) + C

\frac{d}{d x} (\frac{x + 8}{3})

x \to 4 lim (\frac{1}{( x - 4 ) ^{3}})

\int 2 x sec (x) tan (x) d x

\frac{d}{d x} (4 sin^{2} (x) + 9 cos^{2} (x))

What is the integral of 1/(u^2-3) ?
The integral of 1/(u^2-3) is -1/(2sqrt(3))(ln| u/(sqrt(3))+1|-ln| u/(sqrt(3))-1|)+C