What is the integral of (x^2-8x-1)/((x-1)^2(x^2+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-8x-1)/((x-1)^2(x^2+1))

\int \frac{x ^{2} - 8 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{x ^{2} - 8 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2} - 8 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} : \frac{1}{x - 1} - \frac{4}{( x - 1 ) ^{2}} + \frac{- x + 4}{x ^{2} + 1}

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

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{1}{x - 1} d x - \int \frac{4}{( x - 1 ) ^{2}} d x + \int \frac{- x + 4}{x ^{2} + 1} d x

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

\int \frac{4}{( x - 1 ) ^{2}} d x = - \frac{4}{x - 1}

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (2 x + 8)

\frac{d}{d x} ([arctan (3 x)]^{2})

d er i v a t i v e 1680 (2 x + 5)^{4}

y^{^{''}} + y = 5 e^{- 8 x}

n = 1 \sum \infty 2 (\frac{2}{3})^{n - 1}

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