Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 85701 by jagoll last updated on 24/Mar/20

$$\int \frac{\sqrt{3\mathrm{x}-1}}{\sqrt{2\mathrm{x}+1}}\mathrm{dx}$$

Commented by john santu last updated on 24/Mar/20

$$lett=\sqrt{2x+1}\Rightarrow x=\frac{t^2-1}{2}$$$$\int \frac{\sqrt{\frac{3t^2-3}{2}-1}}{t}\times tdt=$$$$\int \sqrt{\frac{3t^2-5}{2}}dt=$$$$\frac{1}{\sqrt{2}}\int \sqrt{3t^2-5}dt=$$$$[\sqrt{3}t=\sqrt{5}\sec u]$$$$\frac{1}{\sqrt{2}}\int \sqrt{5\tan {}^{2}u}\times \frac{\sqrt{5}}{\sqrt{3}}\sec u\tan udu=$$$$\frac{5}{\sqrt{6}}\int \tan {}^{2}u\sec udu=$$$$\frac{5}{\sqrt{6}}[\int \sec {}^{3}udu-\ln \mid \sec u+\tan u\mid +c$$$$\frac{5}{2\sqrt{6}}\sec u\tan u-\frac{5}{2\sqrt{6}}\ln \mid \sec u+\tan u\mid +c$$$$\frac{5}{2\sqrt{6}}.\frac{\sqrt{3}}{\sqrt{5}}t\frac{\sqrt{3t^2-5}}{\sqrt{5}}-\frac{5}{2\sqrt{6}}\ln \mid \frac{\sqrt{3}t+\sqrt{3t^2-5}}{\sqrt{5}}\mid +c$$$$\frac{1}{2\sqrt{2}}\sqrt{2x+1}\sqrt{6x-2}-\frac{5}{2\sqrt{6}}\ln \mid \frac{\sqrt{6x+3}+\sqrt{6x-2}}{\sqrt{5}}\mid +c$$

Commented by mathmax by abdo last updated on 24/Mar/20

$$I=\int \sqrt{\frac{3x-1}{2x+1}}dxchangement\sqrt{\frac{3x-1}{2x+1}}=tgive$$$$\frac{3x-1}{2x+1}=t^2\Rightarrow 3x-1=2t^{2}x+t^2\Rightarrow (3-2t^2)x=t^2+1\Rightarrow$$$$x=\frac{t^2+1}{3-2t^2}\Rightarrow dx=\frac{2t(3-2t^2)-(t^2+1)(-4t)}{{(3-2t^2)}^2}dt$$$$=\frac{6t-4t^3+4t^3+4t}{{(2t^2-3)}^2}dt=\frac{10t}{{(2t^2-3)}^2}dt\Rightarrow I=\int t\times \frac{10t}{{(2t^2-3)}^2}dt$$$$=5\int \frac{2t^2-3+3}{{(2t^2-3)}^2}dt=5\int \frac{dt}{2t^2-3}+15\int \frac{dt}{{(2t^2-3)}^2}wehave$$$$\int \frac{dt}{2t^2-3}=\frac{1}{2}\int \frac{dt}{t^2-{\left(\sqrt{\frac{3}{2}}\right)}^2}=\frac{1}{2\times 2\sqrt{\frac{3}{2}}}\int (\frac{1}{t-\sqrt{\frac{3}{2}}}-\frac{1}{t+\sqrt{\frac{3}{2}}})$$$$=\frac{\sqrt{2}}{4\sqrt{3}}ln\mid \frac{t-\sqrt{\frac{3}{2}}}{t+\sqrt{\frac{3}{2}}}\mid +c_0$$$$\int \frac{dt}{{(2t^2-3)}^2}=\frac{1}{4}\int \frac{dt}{{(t-\sqrt{\frac{3}{2}})}^2{(t+\sqrt{\frac{3}{2}})}^2}$$$$=\frac{a}{t-\sqrt{\frac{3}{2}}}+\frac{b}{{(t-\sqrt{\frac{3}{2}})}^2}+\frac{c}{t+\sqrt{\frac{3}{2}}}+\frac{d}{{(t+\sqrt{\frac{3}{2}})}^2}=F\left(t\right)$$$$\Rightarrow \int F\left(t\right)dt=aln\mid t-\sqrt{\frac{3}{2}}\mid -\frac{b}{t-\sqrt{\frac{3}{2}}}+cln\mid t+\sqrt{\frac{3}{2}}\mid -\frac{d}{t+\sqrt{\frac{3}{2}}}$$$$witht=\sqrt{\frac{3x-1}{2x+1}}restcalculusofcoefficient...becontinued...$$

Commented by john santu last updated on 24/Mar/20

$$wrong?where?$$

Answered by MJS last updated on 24/Mar/20

$$\int \frac{\sqrt{3x-1}}{\sqrt{2x+1}}dx=$$$$[t=\frac{\sqrt{2x+1}}{\sqrt{3x-1}}\to dx=-\frac{2}{5}\sqrt{2x+1}\sqrt{{(3x-1)}^3}dt]$$$$=-\frac{10}{9}\int \frac{dt}{{(t^2-\frac{2}{3})}^2}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{5t}{6(t^2-\frac{2}{3})}+\frac{5}{6}\int \frac{dt}{t^2-\frac{2}{3}}=$$$$=\frac{5t}{6(t^2-\frac{2}{3})}+\frac{5\sqrt{6}}{24}\ln \frac{3t-\sqrt{6}}{3t+\sqrt{6}}=$$$$=\frac{1}{2}\sqrt{3x-1}\sqrt{2x+1}+\frac{5\sqrt{6}}{24}\ln (6x+\frac{1}{2}-\sqrt{3x-1}\sqrt{2x+1})+C$$

Commented by MJS last updated on 24/Mar/20

$$\mathrm{sorry}\mathrm{your}\mathrm{solution}\mathrm{is}\mathrm{ok},\mathrm{I}\mathrm{made}\mathrm{a}\mathrm{mistake}$$$$\mathrm{while}\mathrm{going}\mathrm{through}\mathrm{it}.$$

Commented by john santu last updated on 24/Mar/20

$$theresultsofanintegral$$$$canbedifferentfromone$$$$anotherbecauseofdifferent$$$$method.buttheyaretrueas$$$$asolution$$

Commented by john santu last updated on 24/Mar/20

$$forexample$$$$\int \sin x\cos xdx=\int \sin xd\left(\sin x\right)$$$$=\frac{1}{2}\sin {}^{2}x+c$$$$\int \sin x\cos xdx=\frac{1}{2}\int \sin 2xdx$$$$=-\frac{1}{4}\cos 2x+c$$

Commented by john santu last updated on 24/Mar/20

$$thankyoumister$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com