What is the integral of ((2x-3))/(x^2+4x+3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((2x-3))/(x^2+4x+3)

\int \frac{( 2 x - 3 )}{x ^{2} + 4 x + 3} d x

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

Solution steps

\int \frac{2 x - 3}{x ^{2} + 4 x + 3} d x

Complete the square

x^{2} + 4 x + 3 : (x + 2)^{2} - 1

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

Apply u-substitution

= \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 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

= ln u^{2} - 1 - (- 7 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣))

Substitute back

u = x + 2

= ln (x + 2)^{2} - 1 - (- 7 (\frac{1}{2} ln ∣ x + 2 + 1 ∣ - \frac{1}{2} ln ∣ x + 2 - 1 ∣))

Simplify

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

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

Add a constant to the solution

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

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