Question

\[\int\frac{x^2}{\sqrt{1 - x}} dx\]

Answer 1

Let I = `int x^2/[ sqrt( x - 1) ] `dx
Substituting x - 1 = t and dx = dt, we get,

I = `int  ( t + 1)^2/sqrt t`dx

= `int ( t^2 + 1 +2t )/ sqrtt` dt

= `int ( t^(3/2) + t^(-1/2) + 2t^(-1/2) )`dt

= `2/5t^(5/2) + 2t^(1/2) + 4/3t^(3/2)` + c

= `[ 6t^(5/2) + 30t^(1/2) + 20t^(3/2)]/15`+ c

= `2/15  t^(1/2)( 3t^2 + 15 + 10t )` + c

= `2/15 sqrt( x - 1 )[ 3( x -1 )^2 + 15 + 10( x - 1)]`+ c

= `2/15 sqrt( x - 1 )[ 3( x^2  + 1 - 2x ) + 15 + 10x - 10]`+ c

= `2/15 sqrt( x - 1 )[  3x^2  + 3 - 6x + 15 + 10x - 10]`+ c

= `2/15 sqrt( x - 1 ) [  3x^2 + 4x + 8 ]`+ c