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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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