What is the integral of 1/(7(x^2+1)) ?

The integral of 1/(7(x^2+1)) is 1/7 arctan(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/(7(x^2+1))

\int \frac{1}{7 ( x ^{2} + 1 )} d x

\frac{1}{7} arctan (x) + C

Solution steps

\int \frac{1}{7 ( x ^{2} + 1 )} d x

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{x ^{2} + 1} d x

Use the common integral:

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

= \frac{1}{7} arctan (x)

Add a constant to the solution

= \frac{1}{7} arctan (x) + C

\frac{d}{d x} ((x^{3} + 7)^{6})

\frac{\partial}{\partial x} (x^{2} + y^{2} - 2 x y)

t \to 1 lim (\frac{ln ( t )}{t - 1})

y^{^{'}} + x y = 4 x y^{\frac{7}{2}}

\int_{0}^{\frac{π}{2}} 4 cos^{2} (x) d x

What is the integral of 1/(7(x^2+1)) ?
The integral of 1/(7(x^2+1)) is 1/7 arctan(x)+C