What is the integral of 1/((64x^2+1)^2) ?

The integral of 1/((64x^2+1)^2) is 1/16 (arctan(8x)+1/2 sin(2arctan(8x)))+C

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integral of 1/((64x^2+1)^2)

\int \frac{1}{( 64 x ^{2} + 1 ) ^{2}} d x

\frac{1}{16} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))) + C

Solution steps

\int \frac{1}{( 64 x ^{2} + 1 ) ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{8 ( 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}{8} \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}{8} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{8} \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}{8} \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}{8} \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}{8} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{8} \cdot \frac{1}{2} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)))

Simplify

\frac{1}{8} \cdot \frac{1}{2} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))) : \frac{1}{16} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)))

= \frac{1}{16} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)))

Add a constant to the solution

= \frac{1}{16} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))) + C

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What is the integral of 1/((64x^2+1)^2) ?
The integral of 1/((64x^2+1)^2) is 1/16 (arctan(8x)+1/2 sin(2arctan(8x)))+C