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

The integral of (x+1)/(x^2-2x+2) is 1/2 ln|x^2-2x+2|+2arctan(x-1)+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)/(x^2-2x+2)

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

\frac{1}{2} ln x^{2} - 2 x + 2 + 2 arctan (x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

\frac{u + 2}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{2}{u ^{2} + 1}

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

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

\int \frac{2}{u ^{2} + 1} d u = 2 arctan (u)

= \frac{1}{2} ln u^{2} + 1 + 2 arctan (u)

Substitute back

u = x - 1

= \frac{1}{2} ln (x - 1)^{2} + 1 + 2 arctan (x - 1)

Simplify

\frac{1}{2} ln (x - 1)^{2} + 1 + 2 arctan (x - 1) : \frac{1}{2} ln x^{2} - 2 x + 2 + 2 arctan (x - 1)

= \frac{1}{2} ln x^{2} - 2 x + 2 + 2 arctan (x - 1)

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 2 x + 2 + 2 arctan (x - 1) + C

\int (x + 6) e^{x} d x

\int \frac{2 - 4 x}{( x + 2 ) ( x - 3 )} d x

\int \frac{2 y ^{5} - 8 y ^{3} - 1}{4 y ^{4}} d y

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

\int 7 sin (- 2 x) d x

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