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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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

\int sin^{3} (t) cos^{3} (t) d t

\int \frac{5 x}{( x + 4 ) ( x + 2 )} d x

\int (1 + 3 x) d x

\int (4 x^{2} - 5)^{\frac{1}{2}} d x

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