\(A=\int \frac{x}{\sqrt{x+2}}dx \\ = \int \frac{x+2-2}{\sqrt{x+2}}dx \\ = \int \sqrt{x+2}-2\frac{1}{\sqrt{x+2}}dx \\ = \frac{2}{3}(x+2)^{\frac{3}{2}}-4\sqrt{x+2}+C\)

\(B=\int \frac{sinx+cosx}{\sqrt[3]{1-sin2x}}dx \\ x=\frac{\pi}{4}-u, dx=-du \\ =- \int \frac{sin(\frac{\pi}{4}-u)+cos(\frac{\pi}{4}-u)}{\sqrt[3]{1-sin(\frac{\pi}{2}-2u)}}du \\ = - \int \frac{\frac{1}{\sqrt2}cosu+\frac{1}{\sqrt2}sinu+\frac{1}{\sqrt2}cosu-\frac{1}{\sqrt2}sinu}{\sqrt[3]{1-cos2u}}du \\ = -\int \frac{\frac{2}{\sqrt2}cosu}{\sqrt[3]{1-cos2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{1-cos^2u+sin^2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{2sin^2u}}du \\ v=sinu, dv=cosudu \\ = -\sqrt2 \int \frac{1}{\sqrt[3]{2v^2}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} \int v^{-\frac{2}{3}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} 3v^\frac{1}{3}+C \\ = -\frac{\sqrt2}{\sqrt[3]2} 3\sqrt[3]{sin(\frac{\pi}{4}-x)}+C \)

\(I= \int \frac{sinx-cosx}{(sinx+cosx)^2-4}\ dx \\u=sinx+cosx, du=(cosx-sinx) dx=-(sinx-cosx)dx \\I = -\int \frac{du}{u^2-4} \\ =-\int \frac{\frac{1}{4}}{u-2}+\frac{\frac{1}{4}}{u+2}\ du \\ = -\frac{1}{4}ln(|\frac{sinx+cosx-2}{sinx+cosx+2}|)+C\)

\(I=\int\dfrac{2}{2+5sinxcosx}dx=\int\dfrac{2sec^2x}{2sec^2x+5tanx}dx\\ =\int\dfrac{2sec^2x}{2tan^2x+5tanx+2}dx\)

We substitute :

\(u=tanx,du=sec^2xdx\\ I=\int\dfrac{2}{2u^2+5u+2}du\\ =\int\dfrac{2}{2\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{8}}du\\ =\int\dfrac{1}{\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{16}}du\\ \)

Then, 

\(t=u+\dfrac{5}{4}\\I=\int\dfrac{1}{t^2-\dfrac{9}{16}}dt\\ =\int\dfrac{\dfrac{2}{3}}{t-\dfrac{3}{4}}-\dfrac{\dfrac{2}{3}}{t+\dfrac{3}{4}}dt\)

Finally,

\(I=\dfrac{2}{3}ln\left(\left|\dfrac{t-\dfrac{3}{4}}{t+\dfrac{3}{4}}\right|\right)+C=\dfrac{2}{3}ln\left(\left|\dfrac{tanx+\dfrac{1}{2}}{tanx+2}\right|\right)+C\)