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

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

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integral of 1/(3-x^2)

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( 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}{3} \cdot \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}{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

\int (sin (5 x) sin (x)) d x

1

\int \frac{x}{2 ( x + 1 ) 2 x + 1} d x

\frac{d}{d x} (\frac{x ^{3}}{3} - \frac{3}{x ^{5}} + 4 π)

\frac{d y}{d x} = \frac{( 4 y lo g _{e} ( x ) )}{x}

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