What is the integral of 1/(16x^4+8x^2+1) ?

The integral of 1/(16x^4+8x^2+1) is 1/4 (arctan(2x)+1/2 sin(2arctan(2x)))+C

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integral of 1/(16x^4+8x^2+1)

\int \frac{1}{16 x ^{4} + 8 x ^{2} + 1} d x

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

Solution steps

\int \frac{1}{16 x ^{4} + 8 x ^{2} + 1} d x

Factor

\frac{1}{16 x ^{4} + 8 x ^{2} + 1}

= \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

= \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

What is the integral of 1/(16x^4+8x^2+1) ?
The integral of 1/(16x^4+8x^2+1) is 1/4 (arctan(2x)+1/2 sin(2arctan(2x)))+C