Answer
see below answer
Work Step by Step
a)
let
$x$ = $\frac{3}{2}tanθ$
$dx$ = $\frac{3}{2}{sec^{2}θ}dθ$
and
$\sqrt {4x^{2}+9}$ = $\sqrt {4(\frac{3}{2}tanθ)^{2}+9}$ = $\sqrt {9tan^{2}θ+9}$ = $3secθ$
so
$I$ = ${\int}\frac{dx}{\sqrt {4x^{2}+9}}$ = ${\int}\frac{\frac{3}{2}sec^{2}θ}{3secθ}$ = $\frac{1}{2}\int{secθ}dθ$ =$\frac{1}{2}\ln|secθ+tanθ|+C$
b)
$I$= $\frac{1}{2}\ln|secθ+tanθ|+C$
$x$ = $\frac{3}{2}tanθ$
$tanθ$ = $\frac{2x}{3}$
c)
$I$= $\frac{1}{2}\ln|secθ+tanθ|+C$
$I$= $\frac{1}{2}\ln|\frac{\sqrt {4x^{2}+9}}{3}+\frac{2x}{3}|+C$
$I$= $\frac{1}{2}\ln|\frac{\sqrt {4x^{2}+9}+2x}{3}|+C$
$I$= $\frac{1}{2}\ln|{\sqrt {4x^{2}+9}+2x}|-\frac{1}{2}\ln3+C$
$I$= $\frac{1}{2}\ln|{\sqrt {4x^{2}+9}+2x}|+C$
