Please solve : ∫(ax+b)^1/2 dx

Let I = ∫(ax+b)^1/2 dx


say ax+b = t , again dt/dx= d/dx(ax+b) = d/dx(ax)+d/dx(b)

= a

=> dt=adx....(1)


now,I = ∫(ax+b)^1/2 dx = ∫(t)^1/2dt/a [from (1)]

=1/a ∫t^(1/2)= 1/a { t^(3/2)}/(3/2)+C

= 2/3a.t^3/2+C

= 2/3a.(ax+b)^(3/2)+C Ans.

Use u-substitution here. Let...

u = ax + b
du = adx ==> dx = du/a

∫(ax+b)^1/2 dx = ∫u^(1/2)(du)/a
.....=(1/a)∫u^(1/2)du
.....=(1/a)[(2/3)u^(3/2)] + C
.....=[2/(3a)]*(ax+b)^(3/2) + C

Or...

.....=[2*(ax+b)^(3/2)]/(3a) +C

They're both the same.

Hope that helps.

EDIT: Whoops, forgot the + C. :)

∫√(ax + b) dx

u = ax + b

du/a = dx

1/a*∫√u du

2/(3a) * u^(3/2) + C

2/(3a) * (ax + b)^(3/2) + C

2/3a (ax+b)^3/2 + c