What is the integral of (18x-6)/(sqrt(3x^2-2x-7)) ?

The integral of (18x-6)/(sqrt(3x^2-2x-7)) is 2sqrt(3)sqrt(9x^2-6x-21)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (18x-6)/(sqrt(3x^2-2x-7))

\int \frac{18 x - 6}{3 x ^{2} - 2 x - 7} d x

23 9 x^{2} - 6 x - 21 + C

Solution steps

\int \frac{18 x - 6}{3 x ^{2} - 2 x - 7} d x

Complete the square

3 x^{2} - 2 x - 7 : 3 (x - \frac{1}{3})^{2} - \frac{22}{3}

= \int \frac{18 x - 6}{3 ( x - \frac{1}{3} ) ^{2} - \frac{22}{3}} d x

Apply u-substitution

: \int \frac{18 3 u}{9 u ^{2} - 22} d u

= \int \frac{18 3 u}{9 u ^{2} - 22} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 183 \cdot \int \frac{u}{9 u ^{2} - 22} d u

Apply u-substitution

: \int \frac{1}{18 v} d v

= 183 \cdot \int \frac{1}{18 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 183 \frac{1}{18} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= 183 \frac{1}{18} \cdot 2 v^{\frac{1}{2}}

Substitute back

= 183 \frac{1}{18} \cdot 2 (9 (x - \frac{1}{3})^{2} - 22)^{\frac{1}{2}}

Simplify

183 \frac{1}{18} \cdot 2 (9 (x - \frac{1}{3})^{2} - 22)^{\frac{1}{2}} : 23 9 x^{2} - 6 x - 21

= 23 9 x^{2} - 6 x - 21

Add a constant to the solution

= 23 9 x^{2} - 6 x - 21 + C

\frac{10}{5 s + 1}

\int 4 sin (x) + \frac{2 x ^{5} - x}{x} d x

f (x) = - 3 x^{4} + 7 x^{2} - 4, at x = 1

\frac{\partial}{\partial x} (2 x - 5 y)

\frac{1}{s ^{2} + 0.1 s + 1}

What is the integral of (18x-6)/(sqrt(3x^2-2x-7)) ?
The integral of (18x-6)/(sqrt(3x^2-2x-7)) is 2sqrt(3)sqrt(9x^2-6x-21)+C