What is the integral of (sec^2(x)+8) ?

The integral of (sec^2(x)+8) is tan(x)+8x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sec^2(x)+8)

\int (sec^{2} (x) + 8) d x

tan (x) + 8 x + C

Solution steps

\int sec^{2} (x) + 8 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sec^{2} (x) d x + \int 8 d x

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

\int 8 d x = 8 x

= tan (x) + 8 x

Add a constant to the solution

= tan (x) + 8 x + C

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

\frac{d}{d x} ((2 x^{2} - 4 x)^{2})

\frac{3}{s s ^{2} + 8 ss + 18 s}

\frac{d}{d x} (4 x - 3 x^{2} - y^{2} - (z - 1)^{2})

\frac{3}{x} - e^{4}

What is the integral of (sec^2(x)+8) ?
The integral of (sec^2(x)+8) is tan(x)+8x+C