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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+7)/((x-1)^2(x+1))

\int \frac{2 x + 7}{( x - 1 ) ^{2} ( x + 1 )} d x

- \frac{5}{4} ln ∣ x - 1 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x - 1} - \frac{7}{2 ( x - 1 )} + C

Solution steps

\int \frac{2 x + 7}{( x - 1 ) ^{2} ( x + 1 )} d x

Expand

\frac{2 x + 7}{( x - 1 ) ^{2} ( x + 1 )} : \frac{2 x}{( x - 1 ) ^{2} ( x + 1 )} + \frac{7}{( x - 1 ) ^{2} ( x + 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 x}{( x - 1 ) ^{2} ( x + 1 )} d x + \int \frac{7}{( x - 1 ) ^{2} ( x + 1 )} d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{2} 18 x^{2} 4^{x^{3}} d x

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

x \to 0 - lim (1 + e^{\frac{1}{x}})

3 t^{2} + 15 t + 15

y = (5 x^{2} - 1)^{- 3}

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