What is the integral of (4arctan(x)-x)/(1+x^2) ?

The integral of (4arctan(x)-x)/(1+x^2) is 2arctan^2(x)-1/2 ln|1+x^2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4arctan(x)-x)/(1+x^2)

\int \frac{4 arctan ( x ) - x}{1 + x ^{2}} d x

2 arctan^{2} (x) - \frac{1}{2} ln 1 + x^{2} + C

Solution steps

\int \frac{4 arctan ( x ) - x}{1 + x ^{2}} d x

Expand

\frac{4 arctan ( x ) - x}{1 + x ^{2}} : \frac{4 arctan ( x )}{1 + x ^{2}} - \frac{x}{1 + x ^{2}}

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

\int \frac{4 arctan ( x )}{1 + x ^{2}} d x = 2 arctan^{2} (x)

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

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

Add a constant to the solution

= 2 arctan^{2} (x) - \frac{1}{2} ln 1 + x^{2} + C

\int \frac{5}{e ^{3 x}} d x

\int cos (2 x) - sin (3 x) d x

\int \frac{- 7 ( x - 9 )}{( x + 5 ) ( x - 2 )} d x

\int \frac{- 4}{( 1 - 2 x ^{2} )} d x

\int 11 tan^{5} (x) sec^{4} (x) d x

What is the integral of (4arctan(x)-x)/(1+x^2) ?
The integral of (4arctan(x)-x)/(1+x^2) is 2arctan^2(x)-1/2 ln|1+x^2|+C