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2x-3-dx-x-2-4x-7-




Question Number 20239 by tammi last updated on 24/Aug/17
∫(((2x+3)dx)/( (√(x^2 +4x−7))))
$$\int\frac{\left(\mathrm{2}{x}+\mathrm{3}\right){dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7}}} \\ $$
Answered by $@ty@m last updated on 25/Aug/17
=∫((2x+4−1)/( (√(x^2 +4x−7))))dx  =∫((2x+4)/( (√(x^2 +4x−7))))dx−∫(dx/( (√((x+2)^2 −((√(11)))^2 ))))  =∫(dt/( (√t)))−ln∣(x+2)+(√((x+2)^2 −(√((11)))^2 ))∣ , where t=x^2 +4x−7  =2(√t)−ln∣x+2+(√(x^2 +4x−7))∣+C  =2(√(x^2 +4x−7))−ln∣x+2+(√(x^2 +4x−7))∣+C
$$=\int\frac{\mathrm{2}{x}+\mathrm{4}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7}}}{dx} \\ $$$$=\int\frac{\mathrm{2}{x}+\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7}}}{dx}−\int\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{11}}\right)^{\mathrm{2}} }} \\ $$$$=\int\frac{{dt}}{\:\sqrt{{t}}}−{ln}\mid\left({x}+\mathrm{2}\right)+\sqrt{\left.\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\sqrt{\left(\mathrm{11}\right.}\right)^{\mathrm{2}} }\mid\:,\:{where}\:{t}={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7} \\ $$$$=\mathrm{2}\sqrt{{t}}−{ln}\mid{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7}}\mid+{C} \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7}}−{ln}\mid{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{7}}\mid+{C} \\ $$

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