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

The integral of (2x-5)/(x^2+2x+2) is ln|x^2+2x+2|-7arctan(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 (2x-5)/(x^2+2x+2)

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

ln x^{2} + 2 x + 2 - 7 arctan (x + 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 7}{u ^{2} + 1} d u

= \int \frac{2 u - 7}{u ^{2} + 1} d u

Expand

\frac{2 u - 7}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{7}{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{2 u}{u ^{2} + 1} d u - \int \frac{7}{u ^{2} + 1} d u

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

\int \frac{7}{u ^{2} + 1} d u = 7 arctan (u)

= ln u^{2} + 1 - 7 arctan (u)

Substitute back

u = x + 1

= ln (x + 1)^{2} + 1 - 7 arctan (x + 1)

Expand

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

= ln x^{2} + 2 x + 2 - 7 arctan (x + 1)

Add a constant to the solution

= ln x^{2} + 2 x + 2 - 7 arctan (x + 1) + C

\int \frac{1}{8 x} d x

n = 0 \sum \infty \frac{1}{n} x^{n}

\int sec^{3} (z) d z

\frac{d}{d x} (sin (\frac{π x}{3}))

f (x) = ln (x^{2} + x + 1)

What is the integral of (2x-5)/(x^2+2x+2) ?
The integral of (2x-5)/(x^2+2x+2) is ln|x^2+2x+2|-7arctan(x+1)+C