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

The integral of (2x^2-1)/((4x-1)(x^2+1)) is -7/34 ln|4x-1|+1/17 (6ln|x^2+1|+3arctan(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{7}{34} ln ∣ 4 x - 1 ∣ + \frac{1}{17} (6 ln x^{2} + 1 + 3 arctan (x)) + C

Solution steps

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

Take the partial fraction of

\frac{2 x ^{2} - 1}{( 4 x - 1 ) ( x ^{2} + 1 )} : - \frac{14}{17 ( 4 x - 1 )} + \frac{12 x + 3}{17 ( x ^{2} + 1 )}

= \int - \frac{14}{17 ( 4 x - 1 )} + \frac{12 x + 3}{17 ( 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{14}{17 ( 4 x - 1 )} d x + \int \frac{12 x + 3}{17 ( x ^{2} + 1 )} d x

\int \frac{14}{17 ( 4 x - 1 )} d x = \frac{7}{34} ln ∣ 4 x - 1 ∣

\int \frac{12 x + 3}{17 ( x ^{2} + 1 )} d x = \frac{1}{17} (6 ln x^{2} + 1 + 3 arctan (x))

= - \frac{7}{34} ln ∣ 4 x - 1 ∣ + \frac{1}{17} (6 ln x^{2} + 1 + 3 arctan (x))

Add a constant to the solution

= - \frac{7}{34} ln ∣ 4 x - 1 ∣ + \frac{1}{17} (6 ln x^{2} + 1 + 3 arctan (x)) + C

f (x) = a^{2}

\frac{x ^{2} - 49}{x - 7}

f (x) = 4 x - 3 x^{2}

(y - 2) d y + x d x = 0

y = x + 2, y = \frac{1}{x + 1}, [0, 2]

What is the integral of (2x^2-1)/((4x-1)(x^2+1)) ?
The integral of (2x^2-1)/((4x-1)(x^2+1)) is -7/34 ln|4x-1|+1/17 (6ln|x^2+1|+3arctan(x))+C