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

The integral of (t^2)/(t^4+1) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)) + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{t}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1))

\int \frac{t}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1))

= - \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1))

Add a constant to the solution

= - \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)) + C

\frac{\partial}{\partial y} (6 e^{yz} cos (x yz))

\frac{d}{d x} (1 1 + x)

\frac{d}{d x} (\frac{x - 3}{x ^{2} - 1})

\frac{d}{d x} ((x + y)^{2})

\int cos (\frac{x}{5}) d x