What is the integral of (x+2)/(x^2+2x+3) ?

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

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

\frac{1}{2} ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} ln (x + 1)^{2} + 2 + \frac{1}{2} arctan (\frac{x + 1}{2}) : \frac{1}{2} ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

= \frac{1}{2} ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x + 3 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{13}} d x

\int \frac{3 x ^{b - 1}}{a + x ^{b}} d x

\int ln (x e^{x}) d x

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

\int 2 x^{3} + \frac{5}{x ^{2}} - \frac{7}{x} d x

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