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

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

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

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

Solution steps

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

Complete the square

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

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

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 = x - \frac{1}{3}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{- 2 x ^{2}}{x ^{4} - 1} d x

\int (8 x^{3} y - y^{3}) d x

\int \frac{12 x ^{3} - 3 x ^{2}}{3 x ^{4} - x ^{3}} d x

\int \frac{2 x ^{4} - x - 4}{x ^{3} + 8} d x

\int (4 x^{2} e^{2 x}) d x

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