What is the integral of 1/((1+9x^2)^2) ?

The integral of 1/((1+9x^2)^2) is 1/6 (arctan(3x)+1/2 sin(2arctan(3x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((1+9x^2)^2)

\int \frac{1}{( 1 + 9 x ^{2} ) ^{2}} d x

\frac{1}{6} (arctan (3 x) + \frac{1}{2} sin (2 arctan (3 x))) + C

Solution steps

\int \frac{1}{( 1 + 9 x ^{2} ) ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{( 1 + u ^{2} ) ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= \frac{1}{3} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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}{3} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{3} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} (arctan (3 x) + \frac{1}{2} sin (2 arctan (3 x)))

Simplify

\frac{1}{3} \cdot \frac{1}{2} (arctan (3 x) + \frac{1}{2} sin (2 arctan (3 x))) : \frac{1}{6} (arctan (3 x) + \frac{1}{2} sin (2 arctan (3 x)))

= \frac{1}{6} (arctan (3 x) + \frac{1}{2} sin (2 arctan (3 x)))

Add a constant to the solution

= \frac{1}{6} (arctan (3 x) + \frac{1}{2} sin (2 arctan (3 x))) + C

\int 2 x ln (x^{3}) d x

\int (5 - 4 x) sin (6 - 5 x) d x

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

\int x 2 - 5 x d x

\int (3 x^{4} - 3 x) d x

What is the integral of 1/((1+9x^2)^2) ?
The integral of 1/((1+9x^2)^2) is 1/6 (arctan(3x)+1/2 sin(2arctan(3x)))+C