What is the integral of (x-7)/(x^2+4x-5) ?

The integral of (x-7)/(x^2+4x-5) is 1/2 ln|x^2+4x-5|+3(1/2 ln| 1/3 (x+2)+1|-1/2 ln| 1/3 (x+2)-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x-7)/(x^2+4x-5)

\int \frac{x - 7}{x ^{2} + 4 x - 5} d x

\frac{1}{2} ln x^{2} + 4 x - 5 + 3 (\frac{1}{2} ln \frac{1}{3} (x + 2) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 2) - 1) + C

Solution steps

\int \frac{x - 7}{x ^{2} + 4 x - 5} d x

Complete the square

x^{2} + 4 x - 5 : (x + 2)^{2} - 9

= \int \frac{x - 7}{( x + 2 ) ^{2} - 9} d x

Apply u-substitution

= \int \frac{u - 9}{u ^{2} - 9} d u

Expand

\frac{u - 9}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} - \frac{9}{u ^{2} - 9}

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} - 9} d u - \int \frac{9}{u ^{2} - 9} d u

\int \frac{u}{u ^{2} - 9} d u = \frac{1}{2} ln u^{2} - 9

\int \frac{9}{u ^{2} - 9} d u = - 3 (\frac{1}{2} ln \frac{u}{3} + 1 - \frac{1}{2} ln \frac{u}{3} - 1)

= \frac{1}{2} ln u^{2} - 9 - (- 3 (\frac{1}{2} ln \frac{u}{3} + 1 - \frac{1}{2} ln \frac{u}{3} - 1))

Substitute back

u = x + 2

= \frac{1}{2} ln (x + 2)^{2} - 9 - (- 3 (\frac{1}{2} ln \frac{x + 2}{3} + 1 - \frac{1}{2} ln \frac{x + 2}{3} - 1))

Simplify

\frac{1}{2} ln (x + 2)^{2} - 9 - (- 3 (\frac{1}{2} ln \frac{x + 2}{3} + 1 - \frac{1}{2} ln \frac{x + 2}{3} - 1)) : \frac{1}{2} ln x^{2} + 4 x - 5 + 3 (\frac{1}{2} ln \frac{1}{3} (x + 2) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 2) - 1)

= \frac{1}{2} ln x^{2} + 4 x - 5 + 3 (\frac{1}{2} ln \frac{1}{3} (x + 2) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 2) - 1)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x - 5 + 3 (\frac{1}{2} ln \frac{1}{3} (x + 2) + 1 - \frac{1}{2} ln \frac{1}{3} (x + 2) - 1) + C

\int cos^{5} (t) sin^{2} (t) d t

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

\int x (a - x) \cdot sin (\frac{nπ x}{a}) d x

\int x e^{5 x^{2} + 9} d x

\int \frac{5}{x ( x + 1 ) ^{2}} d x

What is the integral of (x-7)/(x^2+4x-5) ?
The integral of (x-7)/(x^2+4x-5) is 1/2 ln|x^2+4x-5|+3(1/2 ln| 1/3 (x+2)+1|-1/2 ln| 1/3 (x+2)-1|)+C