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ln-1-x-2-2x-4-dx-




Question Number 146959 by Gbenga last updated on 16/Jul/21
∫ln(1+(√(x^2 +2x+4)))dx
$$\int{ln}\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\right){dx} \\ $$
Answered by mindispower last updated on 17/Jul/21
⇔∫ln(1+(√(y^2 +3)))dy  ⇔yln(1+(√(y^2 +3))))−∫((2y^2 )/(1+(√(y^2 +3))))dy=A−B  y=(√3)sh(w)  ⇔B=∫((6(√3)sh^2 (w)ch(w))/(1+(√3)ch(w)))dw,e^w =t  B⇔∫((p(t))/(q(t)))dt,fraction of polynomil easy to find
$$\Leftrightarrow\int{ln}\left(\mathrm{1}+\sqrt{{y}^{\mathrm{2}} +\mathrm{3}}\right){dy} \\ $$$$\left.\Leftrightarrow{yln}\left(\mathrm{1}+\sqrt{{y}^{\mathrm{2}} +\mathrm{3}}\right)\right)−\int\frac{\mathrm{2}{y}^{\mathrm{2}} }{\mathrm{1}+\sqrt{{y}^{\mathrm{2}} +\mathrm{3}}}{dy}={A}−{B} \\ $$$${y}=\sqrt{\mathrm{3}}{sh}\left({w}\right) \\ $$$$\Leftrightarrow{B}=\int\frac{\mathrm{6}\sqrt{\mathrm{3}}{sh}^{\mathrm{2}} \left({w}\right){ch}\left({w}\right)}{\mathrm{1}+\sqrt{\mathrm{3}}{ch}\left({w}\right)}{dw},{e}^{{w}} ={t} \\ $$$${B}\Leftrightarrow\int\frac{{p}\left({t}\right)}{{q}\left({t}\right)}{dt},{fraction}\:{of}\:{polynomil}\:{easy}\:{to}\:{find} \\ $$
Commented by Gbenga last updated on 17/Jul/21
thanks sir
$${thanks}\:{sir}\: \\ $$

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