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

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

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integral of (1+w)/(3-w^2)

\int \frac{1 + w}{3 - w ^{2}} d w

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

Solution steps

\int \frac{1 + w}{3 - w ^{2}} d w

Expand

\frac{1 + w}{3 - w ^{2}} : \frac{1}{3 - w ^{2}} + \frac{w}{3 - w ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{3 - w ^{2}} d w + \int \frac{w}{3 - w ^{2}} d w

\int \frac{1}{3 - w ^{2}} d w = \frac{1}{2 3} (ln \frac{w}{3} + 1 - ln \frac{w}{3} - 1)

\int \frac{w}{3 - w ^{2}} d w = - \frac{1}{2} ln 3 - w^{2}

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

Add a constant to the solution

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

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