What is the integral of (x^2)/((1+x^2)^3) ?

The integral of (x^2)/((1+x^2)^3) is 1/8 (arctan(x)-1/4 sin(4arctan(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)^3)

integral

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

\frac{1}{8} (arctan (x) - \frac{1}{4} sin (4 arctan (x))) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{sec ^{6} ( u ) cos ^{2} ( u )} d u

= \int \frac{tan ^{2} ( u )}{sec ^{6} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

= \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int 1 - cos (4 u) d u

Apply the Sum Rule:

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

= \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arctan (x)

= \frac{1}{8} (arctan (x) - \frac{1}{4} sin (4 arctan (x)))

Add a constant to the solution

= \frac{1}{8} (arctan (x) - \frac{1}{4} sin (4 arctan (x))) + C

sin (2 x^{2})

\frac{4 - 3 x}{( x + 2 ) ( x ^{2} + 1 )}

\frac{( x + 1 ) ^{3}}{x ^{3}}

cot^{4} (x) - cot^{2} (x)

\frac{6}{x ^{3}}

What is the integral of (x^2)/((1+x^2)^3) ?
The integral of (x^2)/((1+x^2)^3) is 1/8 (arctan(x)-1/4 sin(4arctan(x)))+C