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Question-206541




Question Number 206541 by luciferit last updated on 18/Apr/24
Answered by lepuissantcedricjunior last updated on 18/Apr/24
∫((3x+5)/( (√(x^2 −10x+51))))dx=k  k=(3/2)∫((2x−5+((10)/3)+5)/( (√(x^2 −10x+51))))dx  k=3(√(x^2 −10x+51))+((25)/2)∫(dx/( (√((x−5)^2 +26))))  k=3(√(x^2 −10x+51))+((25)/(2(√(26))))∫(dx/( (√(1+(((x−5)/( (√(26)))))^2 ))))  ((x−5)/( (√(26))))=sin(ha)=>dx=(√(26))cos(ha)da  k=3(√(x^2 −10x+51))+((25)/2)argsin(((x−5)/( (√(26)))))+c
$$\int\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{5}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}}\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{5}+\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{5}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{k}}=\mathrm{3}\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}+\frac{\mathrm{25}}{\mathrm{2}}\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\left(\boldsymbol{{x}}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{26}}} \\ $$$$\boldsymbol{{k}}=\mathrm{3}\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}+\frac{\mathrm{25}}{\mathrm{2}\sqrt{\mathrm{26}}}\int\frac{{d}\boldsymbol{{x}}}{\:\sqrt{\mathrm{1}+\left(\frac{\boldsymbol{{x}}−\mathrm{5}}{\:\sqrt{\mathrm{26}}}\right)^{\mathrm{2}} }} \\ $$$$\frac{\boldsymbol{{x}}−\mathrm{5}}{\:\sqrt{\mathrm{26}}}=\boldsymbol{{sin}}\left(\boldsymbol{{ha}}\right)=>\boldsymbol{{dx}}=\sqrt{\mathrm{26}}\boldsymbol{{cos}}\left(\boldsymbol{{ha}}\right)\boldsymbol{{da}} \\ $$$$\boldsymbol{{k}}=\mathrm{3}\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{x}}+\mathrm{51}}+\frac{\mathrm{25}}{\mathrm{2}}\boldsymbol{{argsin}}\left(\frac{\boldsymbol{{x}}−\mathrm{5}}{\:\sqrt{\mathrm{26}}}\right)+\boldsymbol{{c}} \\ $$$$ \\ $$

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