Intergrate the following with respect to x
(i) `3sqrt(x^5)` (ii) `sqrtx - (1)/(sqrtx)` (iii) `(sqrtx + (1)/(x) )^2`

Options

A. (i) `(6)/(7)x^(7/2) +C` (ii) `(2)/(3)x^(3//2) - 2x^(1//2) +C` (iii) `(x)/(2) - (1/x) + 4x^(1/2)`
B. (i) `(6)/(7)x^(7/2) +C` (ii) `(2)/(3)x^(3//2) - 2x^(1//2) +C` (iii) `(x^2)/(2) - (1/x) + 4x^(1/2)`
C. (i) `(3)/(7)x^(7/2) +C` (ii) `(2)/(3)x^(3//2) - 2x^(1//2) +C` (iii) `(x^3)/(2) - (1/x) + 4x^(1/2)`
D. (i) `(6)/(7)x^(1/2) +C` (ii) `(2)/(3)x^(3//2) - 3x^(1//2) +C` (iii) `(x^2)/(2) - (1/x) + 4x^(1/2)`

Answer

B

Explanation

(i) `int 3sqrtx^5 = dx = int x^(5//3) dx=(x^(5//3+1))/((5)/(3)+1) =(x^(8//3))/(8//3) = (3)/(8)x^(8//3) +C`
(ii) `int(sqrtx - (1)/(sqrtx))dx =int(x^(1//2) - x^(-1//2))dx -int x^(1//2) dx - int x^(-1//2) dx=x^((1//2)+1)/((1)/(2) +1) - (x^(-1//2)+1)/(-(1)/(2)+1) +C`
`=(x^(3//2))/(3//2) - x^(1//2)/(1//2)+C = (2)/(3)x^(3//2) -2x^(1//2) +C`
(iii) ` int sqrt(3x -4)^3 dx = int (3x -4)^(3//2) dx = ((3x -4)^((3//2))+1)/(((3)/(2)+1)(d)/(dx) (3x -4) +C`
`=((3x -4)^(5//2))/((5)/(2)xx3) +C = (2)/(15)(3x -4)^(5//2) +C`
(iv) `int (sqrtx+(1)/(sqrtx))^2 dx = int (x +(1)/(2) +2) dx = (x^2)/(2) +log_e x+2x +C`
(v) `int(x +(1)/(x))3 dx = int(x^3 +3x +(3)/(x) +(1)/(x^3)) dx = (x^4)/(4) +(3x^2)/(2) +3log_e x +(x^(-3 +1))/(-3+1)`
`=(x^4)/(4) +(3x^2)/(2) +3 log_e x -(1)/(2x^2) +C`
(iv) `int (3x^2+x^(-2) -2x)dx =3(x^3)/(3)+x^(-2+1)/(-2+1) -2(x^2)/(2) +C = x^3 - (1)/(x) - x^2 +C`