What is the integral of (4x-11)/(sqrt(x^2-4x-5)) ?

The integral of (4x-11)/(sqrt(x^2-4x-5)) is 4sqrt(x^2-4x-5)-3ln(1/3 |sqrt(x^2-4x-5)+x-2|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x-11)/(sqrt(x^2-4x-5))

\int \frac{4 x - 11}{x ^{2} - 4 x - 5} d x

4 x^{2} - 4 x - 5 - 3 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2) + C

Solution steps

\int \frac{4 x - 11}{x ^{2} - 4 x - 5} d x

Expand

\frac{4 x - 11}{x ^{2} - 4 x - 5} : \frac{4 x}{x ^{2} - 4 x - 5} - \frac{11}{x ^{2} - 4 x - 5}

= \int \frac{4 x}{x ^{2} - 4 x - 5} - \frac{11}{x ^{2} - 4 x - 5} 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{4 x}{x ^{2} - 4 x - 5} d x - \int \frac{11}{x ^{2} - 4 x - 5} d x

\int \frac{4 x}{x ^{2} - 4 x - 5} d x = 4 (x^{2} - 4 x - 5 + 2 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2))

\int \frac{11}{x ^{2} - 4 x - 5} d x = 11 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2)

= 4 (x^{2} - 4 x - 5 + 2 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2)) - 11 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2)

Simplify

4 (x^{2} - 4 x - 5 + 2 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2)) - 11 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2) : 4 x^{2} - 4 x - 5 - 3 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2)

= 4 x^{2} - 4 x - 5 - 3 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2)

Add a constant to the solution

= 4 x^{2} - 4 x - 5 - 3 ln (\frac{1}{3} x^{2} - 4 x - 5 + x - 2) + C

\int 8 - 2 x d x

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

\int 3 x + \frac{3}{5} d x

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

\int 12 sec (2 x) tan (2 x) d x

What is the integral of (4x-11)/(sqrt(x^2-4x-5)) ?
The integral of (4x-11)/(sqrt(x^2-4x-5)) is 4sqrt(x^2-4x-5)-3ln(1/3 |sqrt(x^2-4x-5)+x-2|)+C