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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= \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

= \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

\int \frac{3 - 7 t}{( t ^{2} + 12 t + 40 ) ^{2}} d t

\int \frac{6}{x x ^{2} - 1} d x

\int tan^{2} (x) + cot^{2} (x) + 4 d x

\int 4 x^{3} \cdot ln (x) d x

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

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