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

The integral of 1/(x^2+100) is 1/10 arctan(x/(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2+100)

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

\frac{1}{10} arctan (\frac{x}{10}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{10 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{10} arctan (u)

Substitute back

u = \frac{x}{10}

= \frac{1}{10} arctan (\frac{x}{10})

Add a constant to the solution

= \frac{1}{10} arctan (\frac{x}{10}) + C

n = 0 \sum \infty (\frac{1}{10})^{n}

5 cos (\frac{t}{3000})

\int e^{- X} d X

\int 3 + csc (x) \cdot cot (x) d x

f (x) = \frac{1 + x}{1 + e ^{x}}

What is the integral of 1/(x^2+100) ?
The integral of 1/(x^2+100) is 1/10 arctan(x/(10))+C