What is the integral of (3x+2)/(sqrt(x^2-4x+20)) ?

The integral of (3x+2)/(sqrt(x^2-4x+20)) is 3sqrt(x^2-4x+20)+8ln(1/4 |x-2+sqrt(x^2-4x+20)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x+2)/(sqrt(x^2-4x+20))

\int \frac{3 x + 2}{x ^{2} - 4 x + 20} d x

3 x^{2} - 4 x + 20 + 8 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20) + C

Solution steps

\int \frac{3 x + 2}{x ^{2} - 4 x + 20} d x

Expand

\frac{3 x + 2}{x ^{2} - 4 x + 20} : \frac{3 x}{x ^{2} - 4 x + 20} + \frac{2}{x ^{2} - 4 x + 20}

= \int \frac{3 x}{x ^{2} - 4 x + 20} + \frac{2}{x ^{2} - 4 x + 20} d x

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{3 x}{x ^{2} - 4 x + 20} d x + \int \frac{2}{x ^{2} - 4 x + 20} d x

\int \frac{3 x}{x ^{2} - 4 x + 20} d x = 3 (x^{2} - 4 x + 20 + 2 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20))

\int \frac{2}{x ^{2} - 4 x + 20} d x = 2 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20)

= 3 (x^{2} - 4 x + 20 + 2 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20)) + 2 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20)

Simplify

3 (x^{2} - 4 x + 20 + 2 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20)) + 2 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20) : 3 x^{2} - 4 x + 20 + 8 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20)

= 3 x^{2} - 4 x + 20 + 8 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20)

Add a constant to the solution

= 3 x^{2} - 4 x + 20 + 8 ln (\frac{1}{4} x - 2 + x^{2} - 4 x + 20) + C

What is the integral of (3x+2)/(sqrt(x^2-4x+20)) ?
The integral of (3x+2)/(sqrt(x^2-4x+20)) is 3sqrt(x^2-4x+20)+8ln(1/4 |x-2+sqrt(x^2-4x+20)|)+C