Each integrand is a power of a linear argument ax+b; apply the power rule to the argument as if it were x, then divide by the coefficient a.
Step 1. Part (i). Here a=1, n=6, so no correction factor is needed beyond dividing by a=1.
∫(x+5)6dx=7(x+5)7+c.
Step 2. Part (ii). Rewrite as (2−3x)−4; here a=−3, n=−4.
∫(2−3x)−4dx=−31⋅−3(2−3x)−3+c=9(2−3x)−3+c=9(2−3x)31+c.
Step 3. Part (iii). Rewrite as (3x+2)1/2; here a=3, n=21.
∫(3x+2)1/2dx=31⋅3/2(3x+2)3/2+c=31⋅32(3x+2)3/2+c=92(3x+2)3/2+c.
Step 4. Check by differentiating. dxd(7(x+5)7)=(x+5)6 ✓. dxd(9(2−3x)31)=91⋅(−3)(2−3x)−4⋅(−3)=99(2−3x)−4=(2−3x)−4 ✓. dxd(92(3x+2)3/2)=92⋅23(3x+2)1/2⋅3=(3x+2)1/2 ✓.
✓Final answer
(i) 7(x+5)7+c (ii) 9(2−3x)31+c (iii) 92(3x+2)3/2+c