You need to use integration by parts formula such that:

`int udv = uv - int vdu`

Selecting `u = arcsin x`  and `dv = x`  yields:

`u = arcsin x => du = 1/(sqrt(1-x^2))`

`dv = xdx => v = x^2/2`

`int x arcsin x dx = (x^2/2)arcsin x - int x^2/(2sqrt(1 - x^2)) dx`

You need to come up with the following substitution

`sin t = x => cost dt= dx`

`int x^2/(2sqrt(1 - x^2)) dx = int sin^2 t/(2sqrt(1 - sin^2 t)) cos t dt `

`int sin^2 t/(2sqrt(1 - sin^2 t)) cos t dt = int sin^2 t/(2sqrt(cos^2 t)) cos t dt `

`int sin^2 t/(2sqrt(cos^2 t)) cos t dt = int sin^2 t/(2cos t) cos t dt` 

You need to use half angle formula such that:

`sin^2 t = (1 - cos 2t)/2`

`intint sin^2 t/2 dt = int (1 - cos 2t)/4 dt`

Using the linearity of integrals yields:

`int (1 - cos 2t)/4 dt = int1/4 dt- int (cos 2t)/4 dt`

`int (1 - cos 2t)/4 dt = 1/4 t - (1/4)(sin 2t)/2 + c`

Substituting back `arcsin x`  for t yields:

`int x^2/(2sqrt(1 - x^2)) dx = (1/4)(arcsin x- (sin(2arcsin x))/2) + c`

`int x arcsin x dx = (x^2/2)arcsin x - (1/4)(arcsin x- (sin(2arcsin x))/2) + c`

Hence, evaluating the given indefinite integral yields `int x arcsin x dx = (x^2/2)arcsin x - (1/4)(arcsin x- (sin(2arcsin x))/2) + c.`