What is the integral of (100)/(1+x^2) ?

The integral of (100)/(1+x^2) is 100arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (100)/(1+x^2)

\int \frac{100}{1 + x ^{2}} d x

100 arctan (x) + C

Solution steps

\int \frac{100}{1 + x ^{2}} d x

Take the constant out:

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

= 100 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 100 arctan (x)

Add a constant to the solution

= 100 arctan (x) + C

\int \frac{x ^{2} + 2 x}{x ^{2} + 4} d x

\int \frac{64}{x ^{2} + 16} d x

\int \frac{[ ln ( x ) ] ^{2}}{x} d x

\int 2 x^{3} e^{- x} d x

\int \frac{x ^{4}}{7 x ^{5} + 1} d x

What is the integral of (100)/(1+x^2) ?
The integral of (100)/(1+x^2) is 100arctan(x)+C