What is the integral of (4x-3)/((x-2)(x^2+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x-3)/((x-2)(x^2+1))

\int \frac{4 x - 3}{( x - 2 ) ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{4 x - 3}{( x - 2 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

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

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

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

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

Add a constant to the solution

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

\int (12 x + 2) tan (15 x^{2} + 5 x) d x

\int 2^{x} \cdot 3^{x} d x

\int 2 sin (2 x) - 3 e^{x} d x

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

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

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