What is the integral of (x+7)/(x^2+4x+8) ?

The integral of (x+7)/(x^2+4x+8) is 1/2 ln|x^2+4x+8|+5/2 arctan((x+2)/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+7)/(x^2+4x+8)

\int \frac{x + 7}{x ^{2} + 4 x + 8} d x

\frac{1}{2} ln x^{2} + 4 x + 8 + \frac{5}{2} arctan (\frac{x + 2}{2}) + C

Solution steps

\int \frac{x + 7}{x ^{2} + 4 x + 8} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 10 u + 29} d u

= \int \frac{u}{u ^{2} - 10 u + 29} d u

Complete the square

u^{2} - 10 u + 29 : (u - 5)^{2} + 4

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

Apply u-substitution

: \int \frac{v + 5}{v ^{2} + 4} d v

= \int \frac{v + 5}{v ^{2} + 4} d v

Expand

\frac{v + 5}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{5}{v ^{2} + 4}

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{v}{v ^{2} + 4} d v + \int \frac{5}{v ^{2} + 4} d v

\int \frac{v}{v ^{2} + 4} d v = \frac{1}{2} ln v^{2} + 4

\int \frac{5}{v ^{2} + 4} d v = \frac{5}{2} arctan (\frac{v}{2})

= \frac{1}{2} ln v^{2} + 4 + \frac{5}{2} arctan (\frac{v}{2})

Substitute back

= \frac{1}{2} ln (x + 7 - 5)^{2} + 4 + \frac{5}{2} arctan (\frac{x + 7 - 5}{2})

Simplify

\frac{1}{2} ln (x + 7 - 5)^{2} + 4 + \frac{5}{2} arctan (\frac{x + 7 - 5}{2}) : \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{5}{2} arctan (\frac{x + 2}{2})

= \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{5}{2} arctan (\frac{x + 2}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{5}{2} arctan (\frac{x + 2}{2}) + C

6 y^{^{'}} - 3 y = 2 e^{- 2 x}, y (0) = 10

ln (1 + 2 x), 1

\int cos^{3} (\frac{x}{3}) d x

y = x + 2, y = \frac{1}{x + 1}, x = 0, x = 2

\frac{d y}{d t} = y^{1.01}

What is the integral of (x+7)/(x^2+4x+8) ?
The integral of (x+7)/(x^2+4x+8) is 1/2 ln|x^2+4x+8|+5/2 arctan((x+2)/2)+C