let
x-1= sinu
dx= cosu du
∫ dx/[x^2.√(2x-x^2)]
=∫ du/(1+sinu)^2
=∫ (1-sinu)^2/(cosu)^4 du
=∫ [1-2sinu + (sinu)^2 ] /(cosu)^4 du
=∫ [2-2sinu - (cosu)^2 ] /(cosu)^4 du
=∫ 2(secu)^4 du -2∫sinu/(cosu)^4 du -∫ (secu)^2 du
=∫ 2(secu)^4 du +2∫dcosu/(cosu)^4 - tanu
=∫ 2(secu)^2 dtanu -(2/3)[1/(cosu)^3] -tanu
=∫ 2[1+(tanu)^2] dtanu - (2/3)[1/(cosu)^3] -tanu
=2tanu + (2/3)(tanu)^3 -tanu - (2/3)[1/(cosu)^3] + C
=tanu + (2/3)(tanu)^3 -(2/3)[1/(cosu)^3] + C
=(x-1)/√(2x-x^2) +(2/3)[(x-1)/√(2x-x^2)]^3 - (2/3)[√(2x-x^2)]^3 + C
x-1= sinu
dx= cosu du
∫ dx/[x^2.√(2x-x^2)]
=∫ du/(1+sinu)^2
=∫ (1-sinu)^2/(cosu)^4 du
=∫ [1-2sinu + (sinu)^2 ] /(cosu)^4 du
=∫ [2-2sinu - (cosu)^2 ] /(cosu)^4 du
=∫ 2(secu)^4 du -2∫sinu/(cosu)^4 du -∫ (secu)^2 du
=∫ 2(secu)^4 du +2∫dcosu/(cosu)^4 - tanu
=∫ 2(secu)^2 dtanu -(2/3)[1/(cosu)^3] -tanu
=∫ 2[1+(tanu)^2] dtanu - (2/3)[1/(cosu)^3] -tanu
=2tanu + (2/3)(tanu)^3 -tanu - (2/3)[1/(cosu)^3] + C
=tanu + (2/3)(tanu)^3 -(2/3)[1/(cosu)^3] + C
=(x-1)/√(2x-x^2) +(2/3)[(x-1)/√(2x-x^2)]^3 - (2/3)[√(2x-x^2)]^3 + C