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

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

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

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

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

Solution steps

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

Factor

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

\int 1 d u = u

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

= \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arctan (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{4 + s}{s ^{2} + 3 s + 1}

\frac{\partial}{\partial x} (2 x^{3} y^{3} + 3 x + 4 y)

\frac{\partial}{\partial x} (4 x^{3} y^{3} + 4)

f (x) = 4 x^{3} - 42 x^{2} + 146 x + 48

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