What is the integral of 1/(6w-2-9w^2) ?

The integral of 1/(6w-2-9w^2) is (3w-1)/(3(-9w^2+6w-2))-1/3 arctan(3w-1)+(3w-1)/(3(9w^2-6w+2))+C

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integral of 1/(6w-2-9w^2)

\int \frac{1}{6 w - 2 - 9 w ^{2}} d w

\frac{3 w - 1}{3 ( - 9 w ^{2} + 6 w - 2 )} - \frac{1}{3} arctan (3 w - 1) + \frac{3 w - 1}{3 ( 9 w ^{2} - 6 w + 2 )} + C

Solution steps

\int \frac{1}{6 w - 2 - 9 w ^{2}} d w

Complete the square

6 w - 2 - 9 w^{2} : - 9 (w - \frac{1}{3})^{2} - 1

= \int \frac{1}{- 9 ( w - \frac{1}{3} ) ^{2} - 1} d w

Apply u-substitution

= \int \frac{1}{- 9 u ^{2} - 1} d u

Apply Integration By Parts

= \frac{u}{- 9 u ^{2} - 1} - \int \frac{18 u ^{2}}{( - 9 u ^{2} - 1 ) ^{2}} d u

\int \frac{18 u ^{2}}{( - 9 u ^{2} - 1 ) ^{2}} d u = \frac{1}{3} arctan (3 u) - \frac{u}{1 + 9 u ^{2}}

= \frac{u}{- 9 u ^{2} - 1} - (\frac{1}{3} arctan (3 u) - \frac{u}{1 + 9 u ^{2}})

Substitute back

u = w - \frac{1}{3}

= \frac{w - \frac{1}{3}}{- 9 ( w - \frac{1}{3} ) ^{2} - 1} - (\frac{1}{3} arctan (3 (w - \frac{1}{3})) - \frac{w - \frac{1}{3}}{1 + 9 ( w - \frac{1}{3} ) ^{2}})

Simplify

\frac{w - \frac{1}{3}}{- 9 ( w - \frac{1}{3} ) ^{2} - 1} - (\frac{1}{3} arctan (3 (w - \frac{1}{3})) - \frac{w - \frac{1}{3}}{1 + 9 ( w - \frac{1}{3} ) ^{2}}) : \frac{3 w - 1}{3 ( - 9 w ^{2} + 6 w - 2 )} - \frac{1}{3} arctan (3 w - 1) + \frac{3 w - 1}{3 ( 9 w ^{2} - 6 w + 2 )}

= \frac{3 w - 1}{3 ( - 9 w ^{2} + 6 w - 2 )} - \frac{1}{3} arctan (3 w - 1) + \frac{3 w - 1}{3 ( 9 w ^{2} - 6 w + 2 )}

Add a constant to the solution

= \frac{3 w - 1}{3 ( - 9 w ^{2} + 6 w - 2 )} - \frac{1}{3} arctan (3 w - 1) + \frac{3 w - 1}{3 ( 9 w ^{2} - 6 w + 2 )} + C

What is the integral of 1/(6w-2-9w^2) ?
The integral of 1/(6w-2-9w^2) is (3w-1)/(3(-9w^2+6w-2))-1/3 arctan(3w-1)+(3w-1)/(3(9w^2-6w+2))+C