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

The integral of 1/(a^2+v^2) is 1/a arctan(v/a)+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^2+v^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{v}{a}

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

Add a constant to the solution

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

x \to 4 + lim (\frac{2}{x - 4})

\int_{1}^{e^{5}} x^{2} ln (x) d x

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

- 10 x^{2} y + \frac{d y}{d x} = 9 x^{2}

\int 3 θ cos (θ) d θ

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