Evaluate : (i) `int((x-1))/(sqrt(x+4))dx` (ii) `intxsqrt(x+2)dx`
(iii) `f(4x-2)sqrt(x^(2)+x++1)dx` (iv) `int((4x+3))/(sqrt(2x^(2)+3x+1))dx`

Explanation

(i) Put `(x+4)=t^(2)` so that `x=(t^(2)-4)anddx=2tdt`.
`:.int((x-1))/(sqrt(x+4))dx=2int((t^(2)-5))/(t)dt`
`=2intt^(2)dt-10intdt=(2t^(3))/(3)-10t+C`
`=(2)/(3)(x+4)^(3//2)-10(x+4)^(1//2)+C`.
(ii) Put `(x+2)=t^(2)` so that `x=(t^(2)-2)anddx=2tdt`.
`:.intxsqrt(x+2)dx=int(t^(2)-2)2dt=2intt^(4)dt-4f^(2)dt`
`=(2t^(5))/(5)-(4t^(3))/(3)+C=(2(x+2)^(5//2))/(5)-(4(x+2)^(3//2))/(3)+C`.
(iii) Put `(x^(2)+x+1)=t` so that (2x+1)dx=dt.
`:.int(4x+2)(sqrt(x^(2)+x+1))dx=2intsqrt(t)dt`
`=(4)/(3)t^(3//2)+C=(4)/(3)(x^(2)+x+1)^(3//2)+C`.
(iv) Put `(2x^(2)+3x+1)=t` so that (4x+3)dx=dt.
`:.int((4x+3))/(sqrt(2x^(2)+3x+1))dx=int(dt)/(sqrt(t))=2sqrt(t)+C=2sqrt(2x^(2)+3x+1)+C`.