Question

help me to evaluate the following integrals;

Answer 1

\[\int\limits_{}^{}\frac{ (2+\sqrt{x})^2dx }{ \sqrt{x}}\]

Answer 2

You can try a u substitution for this one as well, \[u = 2+\sqrt{x}\]

Answer 3

how about dx?

Answer 4

Differentiate it respect to x, what do you get?

Answer 5

x^-1/2 ?

Answer 6

Careful, we need to apply power rule, \[\sqrt{x} \implies x^{1/2}\]
\[\frac{ d }{ dx } x^n = nx^n\] right?

Answer 7

1/2

Answer 8

You are very close however, \[\frac{ du }{ dx } = \frac{ 1 }{ 2 }x^{-1/2} \implies \frac{ 1 }{ 2\sqrt{x} }\]

Answer 9

\[\int\limits \frac{ (2+\sqrt{x})^2 }{ \sqrt{x} }dx\] is your original integral, we make a u sub here of \[u=2+\sqrt{x}\] so our integral becomes \[\int\limits \frac{ u^2 }{ \sqrt{x} } dx\] but that looks a bit weird since we have two variables right? So we differentiate our u substitution to get \[u = 2 + \sqrt{x} \implies \frac{ du }{ dx } = \frac{ 1 }{ 2\sqrt{x} } \implies du = \frac{ dx }{ 2\sqrt{x} }\] can you finish it off?

Answer 10

(2+sqrt of x )^3 / 3+c

Answer 11

Close, \[\int\limits u^2 \color{red}{\frac{ dx }{ \sqrt{x} }}\] replace with out substitution of \[2du = \color{red}{\frac{ dx }{ \sqrt{x} }}\] then we have \[\int\limits 2u^2 du\] you forgot about that 2 it seems, but everything else would be good, so our final answer would get us \[\frac{ 2 }{ 3 }(\sqrt{x}+2)^3+C\]