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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^2))/(1+x^2)

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

- arctan (x) + x + C

Solution steps

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

\frac{x ^{2}}{1 + x ^{2}} = - \frac{1}{1 + x ^{2}} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d x = x

= - arctan (x) + x

Add a constant to the solution

= - arctan (x) + x + C

y = \frac{4}{5 - 3 ( x )}

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

\int_{0}^{4} x (4 - x) d x

d y - (y - 1)^{2} = d x

f (x) = x^{2} + 9, at x = 4

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