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




Question Number 65307 by divyajyoti last updated on 28/Jul/19
∫(((4x+3)dx)/( (√(2x^2 +2x−3)))) = ?
$$\int\frac{\left(\mathrm{4}{x}+\mathrm{3}\right){dx}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}}\:=\:?\: \\ $$
Answered by divyajyoti last updated on 28/Jul/19
=∫(dt/( (√t)))+(1/( (√2)))∫(dx/( (√((x+(1/2))^2 −(((√7)/2))^2 ))))  =2(√(2x^2 +2x−3))+        (1/( (√2)))ln ∣x+(1/2)+(√(x^2 +x−(3/2)))∣+c .
$$=\int\frac{{dt}}{\:\sqrt{{t}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}+ \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} +{x}−\frac{\mathrm{3}}{\mathrm{2}}}\mid+{c}\:. \\ $$
Commented by divyajyoti last updated on 28/Jul/19
(new user)I solved it myself, hahaha!  just learning how to use the app.
$$\left({new}\:{user}\right){I}\:{solved}\:{it}\:{myself},\:{hahaha}! \\ $$$${just}\:{learning}\:{how}\:{to}\:{use}\:{the}\:{app}. \\ $$

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