What is the integral of 1/(25x^2+1) ?

The integral of 1/(25x^2+1) is 1/5 arctan(5x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(25x^2+1)

\int \frac{1}{25 x ^{2} + 1} d x

\frac{1}{5} arctan (5 x) + C

Solution steps

\int \frac{1}{25 x ^{2} + 1} d x

Apply Integral Substitution

= \int \frac{1}{5 ( 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}{5} \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}{5} arctan (u)

Substitute back

u = 5 x

= \frac{1}{5} arctan (5 x)

Add a constant to the solution

= \frac{1}{5} arctan (5 x) + C

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

\int - 3 x^{2} + 1 x d x

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

\int \frac{x}{x ^{3} - 3 x + 2} d x

\int 5^{x} cos (x) d x

What is the integral of 1/(25x^2+1) ?
The integral of 1/(25x^2+1) is 1/5 arctan(5x)+C