What is the integral of x(arctan^2(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(arctan^2(x))

\int x (arctan^{2} (x)) d x

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

Solution steps

\int x arctan^{2} (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( x ^{2} + 2 x + 3 )}{( x - 1 ) ( x + 1 ) ^{2}} d x

\int \frac{t}{36 + 4 t ^{4}} d t

\int - 3 csc^{2} (x) d x

\int \frac{x + 7}{2 x ^{3} - 18 x} d x

\int \frac{3 x ^{2} - 1}{x ^{2}} d x

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