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

The integral of 1/((4x^2+1)^2) is 1/4 (arctan(2x)+1/2 sin(2arctan(2x)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x))) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x))) : \frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x)))

= \frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x)))

Add a constant to the solution

= \frac{1}{4} (arctan (2 x) + \frac{1}{2} sin (2 arctan (2 x))) + C

x^{2} - 4, [- 4, 3]

\frac{\partial}{\partial x} (x^{2} y + z cos (y))

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

y^{^{'''}} - 6 y^{^{''}} - y^{^{'}} + 6 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 3 e^{- 5 t}

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