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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

integral

\frac{1}{( a + x ^{2} )}

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

Solution steps

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

a + x^{2} = a (1 + \frac{x ^{2}}{a})

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

Take the constant out:

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

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

Apply Integral Substitution

= \frac{1}{a} \cdot \int \frac{a}{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}{a} a \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}{a} a arctan (u)

Substitute back

u = \frac{1}{a} x

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

Simplify

\frac{1}{a} a arctan (\frac{1}{a} x) : \frac{1}{a} arctan (\frac{1}{a} x)

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

Add a constant to the solution

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

\frac{1}{( 1 + sin ( x ) )}

x^{- 1} 0.01

\frac{1}{cot ^{3} ( x )}

\frac{x}{( x + b ) ^{3}}

y = 2 e^{4 x - 3} 6

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