Integral $\int\frac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x} dx$

Since it is your first question I will be nice this time. Next time, please learn how to write mathjax code to make your math look better, and please show some effort or give some context. That will make your questions much better.

In both your integrals, let $$ u=\cos x-\sin x. $$ The first one will transform into $$ \int\Bigl(\frac{2/3}{1+u^2}+\frac{1/3}{u^2-2}-1\Bigr)\,du $$ while the second one will be $$ \int\frac{1-u^2}{u^4-u^2-2}\,du. $$ I'm sure you can take care of these.

$$a^{n+1}+b^{n+1}=(a+b)(a^n+b^n)-ab(a^{n-1}+b^{n-1})$$ implies $$\dfrac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x}=(\sin x+\cos x)-\dfrac{\sin x \cos x}{\sin^3 x + \cos^3 x}$$ and $$\sin^3 x + \cos^3 x=(\sin x+\cos x)(1-\sin x\cos x).$$ Therefore $$-\dfrac{\sin x \cos x}{\sin^3 x + \cos^3 x}=\dfrac{1}{\sin x +\cos x}-\dfrac{1}{\sin^3 x + \cos^3 x}.$$