Looks like the integral should be
[tex]\displaystyle\int\frac{x+2}{\sqrt{x+(2x+3)^2}}\,\mathrm dx[/tex]
Expand the denominator:
[tex]\displaystyle\int\frac{x+2}{\sqrt{4x^2+13x+9}}\,\mathrm dx[/tex]
Notice that
[tex]u=4x^2+13x+9\implies \mathrm du=(8x+13)\,\mathrm dx[/tex]
In order to use this substitution, expand the integral as
[tex]\displaystyle\frac18\int\frac{8x+13}{\sqrt{4x^2+13x+9}}\,\mathrm dx+\frac38\int\frac{\mathrm dx}{\sqrt{4x^2+13x+9}}[/tex]
In the first integral, use the previously mentioned substitution:
[tex]\displaystyle\frac18\int\frac{8x+13}{\sqrt{4x^2+13x+9}}\,\mathrm dx=\frac18\int\frac{\mathrm du}{\sqrt u}=\frac14\sqrt u[/tex]
[tex]=\dfrac14\sqrt{4x^2+13x+9}[/tex]
In the second integral, complete the square in the denominator:
[tex]4x^2+13x+9=4\left(x+\dfrac{13}8\right)^2-\dfrac{25}{16}[/tex]
Now substitute
[tex]x=\dfrac58\sec t-\dfrac{13}8\implies\mathrm dx=\dfrac58\sec t\tan t\,\mathrm dt[/tex]
so that
[tex]\sec t=\dfrac{8x+13}5[/tex]
Then
[tex]4x^2+13x+9=\dfrac{25}{16}(\sec^2t-1)=\dfrac{25}{16}\tan^2t[/tex]
and the integral becomes
[tex]\displaystyle\frac{\frac{15}{64}}{\frac54}\int\frac{\sec t\tan t}{\sqrt{\tan^2t}}\,\mathrm dt=\frac3{16}\int\sec t\,\mathrm dt[/tex]
[tex]=\dfrac3{16}\ln|\sec t+\tan t|[/tex]
[tex]=\dfrac3{16}\ln\left|\dfrac{8x+13}5+\dfrac{\sqrt{(8x+13)^2-25}}5\right|[/tex]
So, the integral is
[tex]\dfrac14\sqrt{4x^2+13x+9}+\dfrac3{16}\ln\left|\dfrac{8x+13}5+\dfrac{\sqrt{(8x+13)^2-25}}5\right|+C[/tex]
