Đi từ bài căn bản said:
[TEX]\red \I^{*}:=\int_{\alpha}^{\beta} \frac{dx}{\sqrt{a x^2+b x+c}}=\frac{1}{\sqrt{a}}\int_{\alpha}^{\beta} \frac{dx}{\sqrt{ x^2+\frac{b}{a} x+\frac{c}{a}}}=\frac{1}{\sqrt{a}}\int_{\alpha}^{ \beta} \frac{d\(x+\frac{b}{2a}+\sqrt{ x^2+\frac{b}{a} x+\frac{c}{a}} \)}{x+\frac{b}{2a}+\sqrt{ x^2+\frac{b}{a} x+\frac{c}{a}}}=\frac{1}{\sqrt{a}}ln\| \frac{\beta+\frac{b}{2a}+\sqrt{ \beta^2+\frac{b}{a}\beta +\frac{c}{a}}}{\alpha+\frac{b}{2a}+\sqrt{ \alpha^2+\frac{b}{a} \alpha+\frac{c}{a}}}\| [/TEX]
[TEX]\I_1:=\int_{\alpha}^{\beta} \frac{\(mx+n\)dx}{\sqrt{a x^2+b x+c}}\ \ \ \ \ \ \(We\ \ have\)\ \ \ mx+n=\frac{m}{2a}\(2ax+b\)+n-\frac{mb}{2a}[/TEX]
[TEX]\ \ \ \ :=\frac{m}{a}\int_{\alpha}^{\beta} \frac{\(2ax+b\) dx}{2\sqrt{a x^2+b x+c}}+\(n-\frac{mb}{2a}\)\int_{\alpha}^{\beta} \frac{dx}{\sqrt{a x^2+b x+c}}[/TEX]
[TEX]\ \ \ \ :=\frac{m}{a}\(\sqrt{a \beta^2+b \beta+c}-\sqrt{a \alpha^2+b \alpha+c}\)+[/TEX][TEX]\red(n-\frac{mb}{2a}\) \I^{*} [/TEX]
[TEX]\I_2:=\int_{\alpha}^{\beta} \frac{dx}{ \(mx+n\)\sqrt{a x^2+b x+c}} \ \ \ \ \ \ \(Setting\)\ \ \ \ mx+n=\frac{1}{t}\righ^{\(form\ \ of \ \ exercise\)}\(\I*\)[/TEX]
[TEX]\I_3:=\int_{\alpha}^{\beta} \frac{\(px+q\)dx}{ \(mx+n\)\sqrt{a x^2+b x+c}} \ \ \ \ \ \ \(We\ \ have\)\ \ \ px+q=\frac{p}{m}\(mx+n\)+q-\frac{pn}{m} [/TEX]
[TEX]\ \ \ \ :=\frac{p}{m}\int_{\alpha}^{\beta} \frac{dx}{\sqrt{a x^2+b x+c}}+\(q-\frac{pn}{m}\)\int_{\alpha}^{\beta} \frac{dx}{\(mx+n\)\sqrt{a x^2+b x+c}}= \(q-\frac{pn}{m}\)\I_2+ \frac{p}{m}\I^*[/TEX]
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