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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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