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

The integral of (t^2-1)/(t^2+1) is -2arctan(t)+t+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 2 arctan (t) + t + C

Solution steps

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

\frac{t ^{2} - 1}{t ^{2} + 1} = - \frac{2}{t ^{2} + 1} + 1

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

\int \frac{2}{t ^{2} + 1} d t = 2 arctan (t)

\int 1 d t = t

= - 2 arctan (t) + t

Add a constant to the solution

= - 2 arctan (t) + t + C

\int 5 - 0.03 x + 0.0004 x^{2} d x

\int \frac{2 x + 6}{x} d x

\int \frac{1}{1 + 2 ^{x}} d x

\int u sec^{2} (u) d u

\int \frac{6 x ^{8}}{sec ( 10 x ^{3} )} d x

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