Question

x^2 square root of 1-x dx

Answer 1

Let, I=∫x2√x−1dx.

Subst. x−1=t2.∴x=t2+1.∴dx=2tdt.

∴I=∫(t2+1)2t⋅2tdt,

=2∫(t4+2t2+1)dt,

=2(t55+2⋅t33+t),

=2t15(3t4+10t2+15),

=215⋅√x−1{3(x−1)2+10(x−1)+15}.

⇒I=215⋅√x−1(3x2+4)