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Question Number 84894 by Power last updated on 17/Mar/20

Commented by abdomathmax last updated on 18/Mar/20

$$I=\int \sqrt{\frac{ax+b}{cx+d}}dxwedothechangement\sqrt{\frac{ax+b}{cx+d}}=t\Rightarrow$$$$\frac{ax+b}{cx+d}=t^2\Rightarrow ax+b={ct}^{2}x+{dt}^2\Rightarrow (a-{ct}^2)x={dt}^2-b\Rightarrow$$$$x=\frac{{dt}^2-b}{a-{ct}^2}\Rightarrow dx=\frac{\left(2dt\right)(a-{ct}^2)-({dt}^2-b)(-2ct)}{{(a-{ct}^2)}^2}dt$$$$=\frac{2adt-2{dct}^3+2{cdt}^3-2bct}{{({ct}^2-a)}^2}=\frac{2(ad-bc)t}{{({ct}^2-a)}^2}\Rightarrow$$$$I=\int t\times \frac{2(ad-bc)t}{{({ct}^2-a)}^2}dt=2(ad-bc)\int \frac{t^2}{{({ct}^2-a)}^2}dt$$$$=\frac{2(ad-bc)}{c}\int \frac{{ct}^2-a+a}{{({ct}^2-a)}^2}dt$$$$=\frac{2(ad-bc)}{c}\int \frac{dt}{{ct}^2-a}+\frac{2a(ad-bc)}{c}\int \frac{dt}{{({ct}^2-a)}^2}$$$$wehave\int \frac{dt}{{ct}^2-a}=\int \frac{dt}{{\left(\sqrt{c}t\right)}^2-{\left(\sqrt{a}\right)}^2}$$$$=\int \frac{dt}{(\sqrt{c}t-\sqrt{a})(\sqrt{c}t+\sqrt{a})}$$$$=\int (\frac{1}{\sqrt{c}t-\sqrt{a}}-\frac{1}{\sqrt{c}t+\sqrt{a}})dt$$$$=\frac{1}{2\sqrt{ac}}ln\mid \frac{\sqrt{c}t-\sqrt{a}}{\sqrt{c}t+\sqrt{a}}\mid +c$$$$\int \frac{dt}{{({ct}^2-a)}^2}=\int \frac{1}{{(\sqrt{c}t-\sqrt{a})}^2{(\sqrt{c}t+\sqrt{a})}^2}$$$$letdecomposeF\left(t\right)=\frac{1}{{(\sqrt{c}t-\sqrt{a})}^2{(\sqrt{c}t+\sqrt{a})}^2}$$$$F\left(t\right)=\frac{\alpha }{\sqrt{c}t-\sqrt{a}}+\frac{\beta }{{(\sqrt{c}t-\sqrt{a})}^2}+\frac{\lambda }{\sqrt{c}t+\sqrt{a}}+\frac{\rho }{{(\sqrt{c}t+\sqrt{a})}^2}$$$$...becontinued...$$

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