Integration of functions with radicals
\(\circ\,\,\,\,\large{F\Big(x,\,\,\sqrt[n]{\frac{\alpha x+\beta}{\gamma x+\delta}}\Big)\,\,\,_{\large{n\in \mathbb{N}}}}\)
$$\text{The substitution is}\,\,t=\sqrt[n]{\frac{\alpha x+\beta}{\gamma x+\delta}}$$
\(\circ\,\,\,\,\Large{\frac{R(x)}{\sqrt{ax^2+bx+c}}}\,\,_{\normalsize{R(x)-\text{rational function}}}\)
$$R(x)=P(x)+\frac{K(x)}{Q(x)}\,\,\,,\,^{P(x)-\text{integer part}}_{\frac{K(x)}{Q(x)}-\text{fractional part}}$$
$$\frac{K(x)}{Q(x)}\,\text{is decomposable into a sum of simple fractions}\Rightarrow$$
$$\Rightarrow \int\frac{R(x)}{\sqrt{ax^2+bx+c}}=\int\frac{P(x)dx}{\sqrt{ax^2+bx+c}}+\sum_{i=1}^{n}\int\frac{A_idx}{(x-\alpha_i)^{k_i}{\sqrt{ax^2+bx+c}}}+\sum_{i=1}^{m}\int\frac{(M_ix+N_i)dx}{(x^2+p_ix+q_i)^{k_i}{\sqrt{ax^2+bx+c}}}$$
$$\text{I.}\int\frac{P(x)dx}{\sqrt{ax^2+bx+c}}= Q(x)\sqrt{ax^2+bx+c}+\gamma\int\frac{dx}{\sqrt{ax^2+bx+c}}\,\,\,\,_{\normalsize{Q(x)'s\,\text{multiplicity}=n-1}}^{\normalsize{P(x)'s\,\text{multiplicity}=n}}\,\,\,(1)$$
$$\gamma\,\,\text{and}\,\,Q(x)\text{'s coffecients are being found by differentiating }\large{(1)}$$
$$\text{II.}\int\frac{Adx}{(x-\alpha)^k\sqrt{ax^2+bx+c}}$$
$$\text{The substitution is}\,\,\,t=\frac{1}{x-\alpha}$$
$$\text{III.}\int\frac{Mx+N}{(x^2+px+q)^m\sqrt{ax^2+bx+c})}dx$$
$$\longrightarrow\text{if }\,\,\,ax^2+bx+c=a(x^2+px+q):$$
$$\int\frac{Mx+N}{(x^2+px+q)^m\sqrt{ax^2+bx+c})}=\int\frac{M_1x+N_1}{(x^2+px+q)^{m+\frac{1}{2}}}dx=\int\frac{\frac{M_1}{2}(2x+p)+N_1-\frac{M_1p}{2}}{(x^2+px+q)^{m+\frac{1}{2}}}dx=$$
$$=C_1\int\frac{d(x^2+px+q)}{(x^2+px+q)^{m+\frac{1}{2}}}+\underbrace{B_1\int\frac{dx}{(x^2+px+q)^{m+\frac{1}{2}}}}_{\text{substitute}\,\,\Large{t=\frac{d}{dx}\sqrt{x^2+px+q}}}$$
$$\longrightarrow\text{if not}:\text{all }x^1\text{ from below have to be removed:}$$
$$\text{(How) The substitution: }\left[ \begin{gathered}
x=\frac{\gamma t+\lambda}{t+1}\,,\,p\ne \frac{b}{a} \\
x=t-\frac{p}{2}\,,\,p=\frac{b}{a}\\
\end{gathered}\right.$$
$$\text{Result: }$$
$$\int\frac{At+B}{(t^2+\lambda)^m\sqrt{\delta t^2+r}}dt=\underbrace{A\int\frac{tdt}{(t^2+\lambda)^m\sqrt{\delta t^2+r}}}_{\text{substitute }\Large{u=\sqrt{\delta t^2+r}} }+\underbrace{B\int\frac{dt}{(t^2+\lambda)^m\sqrt{\delta t^2+r}}}_{\text{substitute }\Large{v=\frac{d}{dx}\sqrt{
\delta t^2+r}}}$$
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