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

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

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

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

2 arctan (2 x) + sin (2 arctan (2 x)) + C

Solution steps

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

Take the constant out:

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

= 8 \cdot \int \frac{1}{( 4 x ^{2} + 1 ) ^{2}} d x

Apply Integral Substitution

= 8 \cdot \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

= 8 \cdot \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

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

Rewrite using trig identities

= 8 \cdot \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

= 8 \cdot \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

= 8 \cdot \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)

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

Substitute back

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

Simplify

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

= 2 arctan (2 x) + sin (2 arctan (2 x))

Add a constant to the solution

= 2 arctan (2 x) + sin (2 arctan (2 x)) + C

\int \frac{6 - 72 x}{1 - 36 x ^{2}} d x

\frac{d}{d x} (\frac{x}{x + 2})

\frac{d}{d x} (\frac{x + 4}{x - 5})

23 x

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

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