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

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

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

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

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

Solution steps

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

Factor

\frac{1}{x ^{4} + 4 x ^{2} + 4}

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

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

\frac{1}{( 2 u ^{2} + 2 ) ^{2}}

= 2 \cdot \int \frac{1}{4 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 2 \frac{1}{4} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 2 \frac{1}{4} \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

= 2 \frac{1}{4} \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

= 2 \frac{1}{4} \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)

= 2 \frac{1}{4} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

ln (w (h - l) + π - T) + ln (l)

\frac{\partial}{\partial x} (x^{2} + 3 x y)

y = \frac{x}{6} - \frac{6}{x}

\frac{\partial}{\partial x} (- 0.1 \cdot cos (6 \cdot x))

x \to 0 lim (\frac{- 11 x ^{2} + 9 x}{x})

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