What is the integral of (t+5)/(t^2+t-2) ?

The integral of (t+5)/(t^2+t-2) is 2ln|t-1|-ln|t+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (t+5)/(t^2+t-2)

\int \frac{t + 5}{t ^{2} + t - 2} d t

2 ln ∣ t - 1 ∣ - ln ∣ t + 2 ∣ + C

Solution steps

\int \frac{t + 5}{t ^{2} + t - 2} d t

Take the partial fraction of

\frac{t + 5}{t ^{2} + t - 2} : \frac{2}{t - 1} - \frac{1}{t + 2}

= \int \frac{2}{t - 1} - \frac{1}{t + 2} d t

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{2}{t - 1} d t - \int \frac{1}{t + 2} d t

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

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

= 2 ln ∣ t - 1 ∣ - ln ∣ t + 2 ∣

Add a constant to the solution

= 2 ln ∣ t - 1 ∣ - ln ∣ t + 2 ∣ + C

What is the integral of (t+5)/(t^2+t-2) ?
The integral of (t+5)/(t^2+t-2) is 2ln|t-1|-ln|t+2|+C