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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{5 ( 1 + u ^{2} )} 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}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} 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

\frac{\partial}{\partial z} (z (1 + e^{x + y}))

\frac{d}{d x} (3 x y^{2} + 2 y cos (x))

\int x^{2} (x^{3} - 1)^{4} d x

\frac{\partial}{\partial x} (\frac{x ^{2} y + 3 z y ^{3}}{x yz})

x y + y^{^{'}} = 80 x

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