Question

integrate the function √(ax + b).dx

Answer 1

Given, $\int{\sqrt{ax+b}}\,dx$
Let us consider (ax + b) = t
differentiate both sides,
a.dx = dt => dx = dt/a , put it above integration

$\int{\sqrt{ax+b}}\,dx=\int{\sqrt{t}}\,\frac{1}{a}dt\\\\=\frac{1}{a}\int{\sqrt{t}}\,dt\\\\=\frac{1}{a}\frac{t^{1/2+1}}{1/2+1}+C\\\\=\frac{2t^{3/2}}{3a}+C$
now, put t = (ax + b)
so, I = $\frac{2(ax+b)^{3/2}}{3a}+C$

Answer 2

HELLO DEAR,


Given, $\bold{ \int{\sqrt{ax+b}}\,dx}$ 

Let (ax + b) = t $\bold{\Rightarrow a.dx = dt }$
$\bold{\Rightarrow dx = dt/a}$


$\bold{\int{\sqrt{ax+b}}\,dx=\int{\sqrt{t}}\,\frac{1}{a}dt}\\\\ \Rightarrow \bold{\frac{1}{a}\int{\sqrt{t}}\,dt} \\\\ \Rightarrow \bold{\frac{1}{a}\frac{t^{1/2+1}}{1/2+1}+C} \\\\ \Rightarrow \bold{\frac{2t^{3/2}}{3a}+C}$

now, put the value of t = (ax + b) in the above function

so, I = $\bold{\frac{2(ax+b)^{3/2}}{3a} + c}$

I HOPE ITS HELP YOU DEAR,
THANKS