What is the integral of (x+1)/(x^2+4x+7) ?

The integral of (x+1)/(x^2+4x+7) is 1/2 ln|x^2+4x+7|-1/(sqrt(3))arctan(1/(sqrt(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)/(x^2+4x+7)

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

\frac{1}{2} ln x^{2} + 4 x + 7 - \frac{1}{3} arctan (\frac{1}{3} (x + 2)) + C

Solution steps

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

Complete the square

x^{2} + 4 x + 7 : (x + 2)^{2} + 3

= \int \frac{x + 1}{( x + 2 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 3} d u

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

Expand

\frac{u - 1}{u ^{2} + 3} : \frac{u}{u ^{2} + 3} - \frac{1}{u ^{2} + 3}

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} + 3} d u - \int \frac{1}{u ^{2} + 3} d u

\int \frac{u}{u ^{2} + 3} d u = \frac{1}{2} ln u^{2} + 3

\int \frac{1}{u ^{2} + 3} d u = \frac{1}{3} arctan (\frac{u}{3})

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

Substitute back

u = x + 2

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

Simplify

\frac{1}{2} ln (x + 2)^{2} + 3 - \frac{1}{3} arctan (\frac{x + 2}{3}) : \frac{1}{2} ln x^{2} + 4 x + 7 - \frac{1}{3} arctan (\frac{1}{3} (x + 2))

= \frac{1}{2} ln x^{2} + 4 x + 7 - \frac{1}{3} arctan (\frac{1}{3} (x + 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 7 - \frac{1}{3} arctan (\frac{1}{3} (x + 2)) + C

y = ln (\frac{e ^{x} + 1}{e ^{x} - 1})

h (v) = 4 v^{2} - 2

f (x) = 9 x^{2} + 7 x - 6, at x = - \frac{7}{18}

\frac{d}{d x} (1 - ∣ x ∣)

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

What is the integral of (x+1)/(x^2+4x+7) ?
The integral of (x+1)/(x^2+4x+7) is 1/2 ln|x^2+4x+7|-1/(sqrt(3))arctan(1/(sqrt(3))(x+2))+C