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

The integral of (4t^2+3t-1)/(t^3-t^2) is 6ln|t-1|-2ln|t|-1/t+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

6 ln ∣ t - 1 ∣ - 2 ln ∣ t ∣ - \frac{1}{t} + C

Solution steps

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

Expand

\frac{4 t ^{2} + 3 t - 1}{t ^{3} - t ^{2}} : \frac{4 t ^{2}}{t ^{3} - t ^{2}} + \frac{3 t}{t ^{3} - t ^{2}} - \frac{1}{t ^{3} - t ^{2}}

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

\int \frac{4 t ^{2}}{t ^{3} - t ^{2}} d t = 4 ln ∣ t - 1 ∣

\int \frac{3 t}{t ^{3} - t ^{2}} d t = 3 (- ln ∣ t ∣ + ln ∣ t - 1 ∣)

\int \frac{1}{t ^{3} - t ^{2}} d t = - ln ∣ t ∣ + \frac{1}{t} + ln ∣ t - 1 ∣

= 4 ln ∣ t - 1 ∣ + 3 (- ln ∣ t ∣ + ln ∣ t - 1 ∣) - (- ln ∣ t ∣ + \frac{1}{t} + ln ∣ t - 1 ∣)

Simplify

4 ln ∣ t - 1 ∣ + 3 (- ln ∣ t ∣ + ln ∣ t - 1 ∣) - (- ln ∣ t ∣ + \frac{1}{t} + ln ∣ t - 1 ∣) : 6 ln ∣ t - 1 ∣ - 2 ln ∣ t ∣ - \frac{1}{t}

= 6 ln ∣ t - 1 ∣ - 2 ln ∣ t ∣ - \frac{1}{t}

Add a constant to the solution

= 6 ln ∣ t - 1 ∣ - 2 ln ∣ t ∣ - \frac{1}{t} + C

What is the integral of (4t^2+3t-1)/(t^3-t^2) ?
The integral of (4t^2+3t-1)/(t^3-t^2) is 6ln|t-1|-2ln|t|-1/t+C