What is the integral of (2t^2+sec^2(t)) ?

The integral of (2t^2+sec^2(t)) is (2t^3)/3+tan(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2t^2+sec^2(t))

\int (2 t^{2} + sec^{2} (t)) d t

\frac{2 t ^{3}}{3} + tan (t) + C

Solution steps

\int 2 t^{2} + sec^{2} (t) 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^{2} d t + \int sec^{2} (t) d t

\int 2 t^{2} d t = \frac{2 t ^{3}}{3}

\int sec^{2} (t) d t = tan (t)

= \frac{2 t ^{3}}{3} + tan (t)

Add a constant to the solution

= \frac{2 t ^{3}}{3} + tan (t) + C

What is the integral of (2t^2+sec^2(t)) ?
The integral of (2t^2+sec^2(t)) is (2t^3)/3+tan(t)+C