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

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

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

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

Solution steps

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

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Substitute back

u = \frac{1}{m} x

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

Simplify

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

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

Add a constant to the solution

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

\int 5 t d t

\int \frac{sin ( x )}{4 + cos ^{2} ( x )} d x

\int y ln (x) + x ln (y) + z^{2} d z

\int sin (32 x) sin (cos (32 x)) d x

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

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