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

The integral of (2t^3-t^2+2t+4)/(1+t^2) is t^2-t+5arctan(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

t^{2} - t + 5 arctan (t) + C

Solution steps

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

Long division

\frac{2 t ^{3} - t ^{2} + 2 t + 4}{1 + t ^{2}} : 2 t - 1 + \frac{5}{t ^{2} + 1}

= \int 2 t - 1 + \frac{5}{t ^{2} + 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 2 t d t - \int 1 d t + \int \frac{5}{t ^{2} + 1} d t

\int 2 t d t = t^{2}

\int 1 d t = t

\int \frac{5}{t ^{2} + 1} d t = 5 arctan (t)

= t^{2} - t + 5 arctan (t)

Add a constant to the solution

= t^{2} - t + 5 arctan (t) + C

\int arccos (x) d x

\int (4 x^{3} - 4 x + 6) d x

\int x^{2} - 6 x + 9 d x

\int - x \cdot cos (x) d x

\int \frac{8 x - 1}{( x - 2 ) ( x + 3 )} d x

What is the integral of (2t^3-t^2+2t+4)/(1+t^2) ?
The integral of (2t^3-t^2+2t+4)/(1+t^2) is t^2-t+5arctan(t)+C