What is the integral of (6x+15)/(x^2+4x+20) ?

The integral of (6x+15)/(x^2+4x+20) is 3ln|x^2+4x+20|+3/4 arctan(1/4 (x+2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (6x+15)/(x^2+4x+20)

\int \frac{6 x + 15}{x ^{2} + 4 x + 20} d x

3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2)) + C

Solution steps

\int \frac{6 x + 15}{x ^{2} + 4 x + 20} d x

Complete the square

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int \frac{6 x + 15}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{6 u + 3}{u ^{2} + 16} d u

Expand

\frac{6 u + 3}{u ^{2} + 16} : \frac{6 u}{u ^{2} + 16} + \frac{3}{u ^{2} + 16}

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{6 u}{u ^{2} + 16} d u + \int \frac{3}{u ^{2} + 16} d u

\int \frac{6 u}{u ^{2} + 16} d u = 3 ln u^{2} + 16

\int \frac{3}{u ^{2} + 16} d u = \frac{3}{4} arctan (\frac{u}{4})

= 3 ln u^{2} + 16 + \frac{3}{4} arctan (\frac{u}{4})

Substitute back

u = x + 2

= 3 ln (x + 2)^{2} + 16 + \frac{3}{4} arctan (\frac{x + 2}{4})

Simplify

3 ln (x + 2)^{2} + 16 + \frac{3}{4} arctan (\frac{x + 2}{4}) : 3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2))

= 3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2))

Add a constant to the solution

= 3 ln x^{2} + 4 x + 20 + \frac{3}{4} arctan (\frac{1}{4} (x + 2)) + C

\int (10 x + 5) d x

\int \frac{x}{x ^{2} + 2 x + 17} d x

\int \frac{( - 1 ) ^{n + 1} ( x - 7 ) ^{n}}{n} d x

\int e^{- in t} d t

\int 9 x \cdot e^{3 x} d x

What is the integral of (6x+15)/(x^2+4x+20) ?
The integral of (6x+15)/(x^2+4x+20) is 3ln|x^2+4x+20|+3/4 arctan(1/4 (x+2))+C