What is the integral of 1/((1+4x^2)(arctan(2x))^3) ?

The integral of 1/((1+4x^2)(arctan(2x))^3) is -1/(4arctan^2(2x))+C

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integral of 1/((1+4x^2)(arctan(2x))^3)

\int \frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}} d x

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

Solution steps

\int \frac{1}{( 1 + 4 x ^{2} ) ( arctan ( 2 x ) ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 arctan ^{3} ( u ) ( 1 + u ^{2} )} d u

= \int \frac{1}{2 arctan ^{3} ( u ) ( 1 + 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 \frac{1}{arctan ^{3} ( u ) ( 1 + u ^{2} )} d u

Apply u-substitution

: \int \frac{1}{v ^{3}} d v

= \frac{1}{2} \cdot \int \frac{1}{v ^{3}} d v

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= \frac{1}{2} (- \frac{1}{2 v ^{2}})

Substitute back

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

Simplify

\frac{1}{2} (- \frac{1}{2 arctan ^{2} ( 2 x )}) : - \frac{1}{4 arctan ^{2} ( 2 x )}

= - \frac{1}{4 arctan ^{2} ( 2 x )}

Add a constant to the solution

= - \frac{1}{4 arctan ^{2} ( 2 x )} + C

\int sin^{4} (6 x) d x

x \to 0 + lim (3 x^{x})

2 x^{2} y^{^{''}} + 6 x y^{^{'}} + 2 y = 0, y (1) = 2, y^{^{'}} (1) = - 1

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 t e^{3 t}

x \to 2 - lim (\frac{4}{( x - 2 ) ^{5}})

What is the integral of 1/((1+4x^2)(arctan(2x))^3) ?
The integral of 1/((1+4x^2)(arctan(2x))^3) is -1/(4arctan^2(2x))+C