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

The integral of (x-2)/((x^2-4x+5)^2) is -1/(2(x^2-4x+5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-2)/((x^2-4x+5)^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} - 4 x + 5

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

Simplify

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

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

Add a constant to the solution

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

(cos (4 x))^{^{'}}

\int_{- 2}^{3} \frac{48}{x ^{4}} d x

\int_{0}^{4 π} t^{2} + 5 d t

\frac{\partial}{\partial x} (x sin (3 y) - z e^{- 2 x})

What is the integral of (x-2)/((x^2-4x+5)^2) ?
The integral of (x-2)/((x^2-4x+5)^2) is -1/(2(x^2-4x+5))+C