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

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

\int \frac{1}{1 + t ^{2}} d t

arctan (t) + C

Solution steps

\int \frac{1}{1 + t ^{2}} d t

Use the common integral:

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

= arctan (t)

Add a constant to the solution

= arctan (t) + C

\int_{1}^{x} 2 (t - 1) d t

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x^{\frac{5}{2}}

\frac{\partial}{\partial x} (7 ln (\frac{3 x y}{7 z}))

\int \frac{sin ^{2} ( x )}{1 - cos ( x )} d x

\frac{d}{d x} (arcsin (9 x^{5}))