Answer:
[tex]2(x-2)^{\frac{3}{2}}+12\sqrt{x-2}+C[/tex]
Step-by-step explanation:
Given:
The expression to integrate is given as:
[tex]\int \frac{3x}{\sqrt{x-2}}\ dx[/tex]
Now, in order to integrate it, we apply the method of substitution.
Let [tex]x-2=t^2[/tex]
[tex]x=t^2+2[/tex]
Differentiating with respect to 't' on both sides, we get:
[tex]\frac{dx}{dt}=2t+0\\\frac{dx}{dt}=2t\\dx=2tdt[/tex]
Replace [tex]dx[/tex] by [tex]2tdt[/tex], [tex]x-2\ by\ t^2[/tex] and [tex]x[/tex] by [tex]t^2+2[/tex]. This gives,
[tex]\int \frac{3(t^2+2)}{\sqrt{t^2}}\ 2tdt\\\\6\int \frac{(t^2+2)t}{t}\ dt\\\\6\int (t^2+2)\ dt\\\\6[\frac{t^3}{3}+2t]+C\\\\2t^3+12t+C[/tex]
Replacing 't' by [tex]\sqrt{x-2}[/tex], we get:
[tex]=2(x-2)\sqrt{x-2}+12\sqrt{x-2}+C[/tex]
[tex]2(x-2)^{\frac{3}{2}}+12\sqrt{x-2}+C[/tex]
Therefore, the integral is:
[tex]\int \frac{3x}{\sqrt{x-2}}\ dx=2(x-2)^{\frac{3}{2}}+12\sqrt{x-2}+C[/tex]
Where 'C' is the constant of integration.
