A child's play!

Let the integral be $\displaystyle I=\int_{0}^{1} \frac{2-3x}{\sqrt{1+3x}} dx = \int_{0}^{1} \frac{3-(1+3x)}{\sqrt{1+3x}}dx = \sqrt{3}\int_{0}^{1}\frac{1}{\sqrt{x+\frac{1}{3}}} dx - \sqrt{3}\int_{0}^{1} \sqrt{x+\frac{1}{3}}dx = 2\sqrt{3}[\sqrt{x+\frac{1}{3}}]_{0}^{1} - \frac{2}{\sqrt{3}}[(x+\frac{1}{3})^{\frac{3}{2}}]_{0}^{1} =2-\frac{14}{9} =\boxed{\frac{4}{9}}$

Let,

$\begin{aligned} I & = \displaystyle \int_{0}^{1}{\dfrac{2-3x}{\sqrt{1+3x}}} \,dx \\ \\ & = \displaystyle \int_{0}^{1}{\dfrac{3-(1+3x)}{\sqrt{1+3x}}} \,dx \\ \\ & = \displaystyle \int_{0}^{1}{\dfrac{3}{\sqrt{1+3x}}} \,dx - \displaystyle \int_{0}^{1}{\sqrt{1+3x}} \,dx \end{aligned}$

Now, let

$1+3x = u \implies 3\,dx = du$

Upon manipulating $I$ to:

$I = \displaystyle \int_{0}^{1}{\dfrac{3}{\sqrt{1+3x}}} \,dx - \dfrac{1}{3} \displaystyle \int_{0}^{1}{3\sqrt{1+3x}} \,dx$

And substituting $3\,dx$ with $du$ , we get,

$I = \displaystyle \int{\dfrac{1}{\sqrt{u}}} \,du - \dfrac{1}{3} \displaystyle \int{\sqrt{u}} \,du$

Now,

$1+3x=u \iff x=\dfrac{u-1}{3}$

The original limits of the problem were $x=0$ to $x=1$ .

Upon substitution, our new limits become $u=1$ to $u=4$ .

Hence,

$\begin{aligned} I & = \displaystyle \int_{1}^{4}{\dfrac{1}{\sqrt{u}}} \,du - \dfrac{1}{3} \displaystyle \int_{1}^{4}{\sqrt{u}} \,du \\ \\ & = {\left(2\sqrt{u} - \dfrac{1}{3}\cdot\dfrac{2}{3}u^{{3}/{2}}\right) \huge |}_{1}^{4} \\ \\ & = {\left(2\sqrt{u} - \dfrac{2}{9}u^{{3}/{2}}\right) \huge |}_{1}^{4} \\ \\ & = \left[ 2\sqrt{4} - \dfrac{2}{9}(4^{{3}/{2}})\right] - \left[ 2\sqrt{1} - \dfrac{2}{9}(1^{{3}/{2}})\right] \\ \\ & = \left(4-\dfrac{16}{9}\right) - \left(2-\dfrac{2}{9}\right) \\ \\ & = 2-\dfrac{14}{9} \\ \\ & = \boxed{\dfrac{4}{9}} \end{aligned}$