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

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

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

x^{2} - 4 x + 3 + ln x^{2} - 4 x + 3 + x - 2 + C

Solution steps

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

Expand

(x - 2)^{2} - 1 : x^{2} - 4 x + 3

= \int \frac{x - 1}{x ^{2} - 4 x + 3} d x

Expand

\frac{x - 1}{x ^{2} - 4 x + 3} : \frac{x}{x ^{2} - 4 x + 3} - \frac{1}{x ^{2} - 4 x + 3}

= \int \frac{x}{x ^{2} - 4 x + 3} - \frac{1}{x ^{2} - 4 x + 3} 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{x}{x ^{2} - 4 x + 3} d x - \int \frac{1}{x ^{2} - 4 x + 3} d x

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

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

= x^{2} - 4 x + 3 + 2 ln x^{2} - 4 x + 3 + x - 2 - ln x^{2} - 4 x + 3 + x - 2

Simplify

= x^{2} - 4 x + 3 + ln x^{2} - 4 x + 3 + x - 2

Add a constant to the solution

= x^{2} - 4 x + 3 + ln x^{2} - 4 x + 3 + x - 2 + C

\int x^{3} - x + 1 d x

\int θ^{2} sec^{2} (θ^{3}) d θ

\int (\frac{x}{2})^{\frac{2}{3}} d x

\int \frac{2 x + 1}{4 x ^{2} + 12 x - 7} d x

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

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