What is the integral of-1/(1+u^2) ?

The integral of-1/(1+u^2) is -arctan(u)+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+u^2)

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

- arctan (u) + C

Solution steps

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

Take the constant out:

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

= - \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= - arctan (u)

Add a constant to the solution

= - arctan (u) + C

3 e^{- 3 s}

x^{2} y^{^{'}} + 2 x y = 19 cos^{2} (x)

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 1, y^{^{'}} (1) = 0

\int r 16 - r^{2} d r

\int_{0}^{2} (3 x^{2} + 2) d x