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

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

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

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

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

Solution steps

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply u-substitution

= \int \frac{u tan ( u )}{sec ^{2} ( u )} d u

Rewrite using trig identities

= \int \frac{1}{2} u sin (2 u) 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 u sin (2 u) d u

Apply Integration By Parts

= \frac{1}{2} (- \frac{1}{2} u cos (2 u) - \int - \frac{1}{2} cos (2 u) d u)

\int - \frac{1}{2} cos (2 u) d u = - \frac{1}{4} sin (2 u)

= \frac{1}{2} (- \frac{1}{2} u cos (2 u) - (- \frac{1}{4} sin (2 u)))

Substitute back

u = arctan (x)

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

Simplify

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

Add a constant to the solution

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

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