What is the integral of (6x^2-5)/(x-1) ?

The integral of (6x^2-5)/(x-1) is 3x^2+6x+ln|x-1|-9+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^2-5)/(x-1)

\int \frac{6 x ^{2} - 5}{x - 1} d x

3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9 + C

Solution steps

\int \frac{6 x ^{2} - 5}{x - 1} d x

Expand

\frac{6 x ^{2} - 5}{x - 1} : \frac{6 x ^{2}}{x - 1} - \frac{5}{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{6 x ^{2}}{x - 1} d x - \int \frac{5}{x - 1} d x

\int \frac{6 x ^{2}}{x - 1} d x = 6 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣)

\int \frac{5}{x - 1} d x = 5 ln ∣ x - 1 ∣

= 6 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣

Simplify

6 (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣) - 5 ln ∣ x - 1 ∣ : 3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9

= 3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9

Add a constant to the solution

= 3 x^{2} + 6 x + ln ∣ x - 1 ∣ - 9 + C

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

\int \frac{1}{36 - 25 x ^{2}} d x

\int 3 sec^{4} (3 x) d x

\int \frac{3 z}{x ^{2} + z ^{2}} d x

What is the integral of (6x^2-5)/(x-1) ?
The integral of (6x^2-5)/(x-1) is 3x^2+6x+ln|x-1|-9+C