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

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

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integral of 1/((x^2-3))

integral

\frac{1}{( x ^{2} - 3 )}

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

Solution steps

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

Apply Integral Substitution

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

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}{u ^{2} - 1} d u

Factor

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

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

Take the constant out:

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

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

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

lo g_{10} (x^{2} + y^{2})

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

\frac{3}{( x ^{2} + 1 )}

1 3^{x} + 1

arctan (x + 1)

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