What is the integral of 8/(1+t^2) ?

The integral of 8/(1+t^2) is 8arctan(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(1+t^2)

\int \frac{8}{1 + t ^{2}} d t

8 arctan (t) + C

Solution steps

\int \frac{8}{1 + t ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{1 + t ^{2}} d t

Use the common integral:

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

= 8 arctan (t)

Add a constant to the solution

= 8 arctan (t) + C

\int \frac{1}{11 x - 7} d x

\int cos (3 x) sin (4 x) d x

\int sin (x) + 3 cos (x) d x

\int x + x^{11} + x^{- 11} d x

\int \frac{csc ( θ )}{4 csc ( θ ) - 3 cot ( θ ) + 1} d θ