What is the integral of (xarctan(x))/((1+x^2)^{3/2)} ?

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

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integral of (xarctan(x))/((1+x^2)^{3/2)}

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

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

Solution steps

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

Apply u-substitution

: \int u tan (u) cos (u) d u

= \int u tan (u) cos (u) d u

Rewrite using trig identities

= \int u sin (u) d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= - u cos (u) - \int - cos (u) d u

\int - cos (u) d u = - sin (u)

= - u cos (u) - (- sin (u))

Substitute back

u = arctan (x)

= - arctan (x) cos (arctan (x)) - (- sin (arctan (x)))

Simplify

- arctan (x) cos (arctan (x)) - (- sin (arctan (x))) : - \frac{arctan ( x )}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

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