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

The integral of ((x+1)^2)/(x^2+1) is ln|x^2+1|+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

ln x^{2} + 1 + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= ln x^{2} + 1 + x

Add a constant to the solution

= ln x^{2} + 1 + x + C

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