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Question Number 2334 by Syaka last updated on 16/Nov/15

∫(dx/((x^2 + 3x + 9)(√(x^2 + 5x + 7)) )) = ?

$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{9}\right)\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\:\mathrm{7}}\:}\:\:=\:? \\ $$

Commented by prakash jain last updated on 16/Nov/15

x^2 +5x+7=(x+(5/2))^2 +(3/4) x+(5/2)=((√3)/2)isin u dx=((√3)/2)icos u du (√((x+(5/2))^2 +(3/4) ))=(√(−(3/4)sin^2 u+(3/4))) =((√3)/2)cos u x^2 +3x+9=(x+(5/2))^2 −2x+((11)/4) =−(3/4)sin^2 u−2(((√3)/2)isin u−(5/2))+((11)/4) =−(1/4)(3sin^2 u+4i(√3)sin u−31) Integral to be computed −(i/4)∫ (du/((3sin^2 u+4(√3)isin u−31))) =((−i)/4)∫ (du/((3sin^2 u+4(√3)isin u+(2i)^2 )−27)) =((−i)/4)∫ (du/(((√3)sin u+2i)^2 −(3(√3))^2 )) =((−i)/4)∫(du/(((√3)sin u+2i−3(√3))((√3)sin u+2k+3(√3)))) =((−i)/(4∙6(√3)))[∫(du/(((√3)sin u+2i−3(√3))))−∫(du/((√3)sin u+2i+3(√3)))] =((−i)/(4∙6∙3))[∫(du/((sin u+((2/(√3))i−3)))−∫(du/(sin u+((2/(√3))i+3)))]

$${x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${x}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\mathrm{sin}\:{u} \\ $$$${dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\mathrm{cos}\:{u}\:{du} \\ $$$$\sqrt{\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:}=\sqrt{−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \:{u}+\frac{\mathrm{3}}{\mathrm{4}}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:{u} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{9}=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} {u}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\mathrm{sin}\:{u}−\frac{\mathrm{5}}{\mathrm{2}}\right)+\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3sin}^{\mathrm{2}} {u}+\mathrm{4}{i}\sqrt{\mathrm{3}}\mathrm{sin}\:{u}−\mathrm{31}\right) \\ $$$$\mathrm{Integral}\:\mathrm{to}\:\mathrm{be}\:\mathrm{computed} \\ $$$$−\frac{{i}}{\mathrm{4}}\int\:\frac{{du}}{\left(\mathrm{3sin}^{\mathrm{2}} {u}+\mathrm{4}\sqrt{\mathrm{3}}{i}\mathrm{sin}\:{u}−\mathrm{31}\right)} \\ $$$$=\frac{−{i}}{\mathrm{4}}\int\:\frac{{du}}{\left(\mathrm{3sin}^{\mathrm{2}} {u}+\mathrm{4}\sqrt{\mathrm{3}}{i}\mathrm{sin}\:{u}+\left(\mathrm{2}{i}\right)^{\mathrm{2}} \right)−\mathrm{27}} \\ $$$$=\frac{−{i}}{\mathrm{4}}\int\:\frac{{du}}{\left(\sqrt{\mathrm{3}}\mathrm{sin}\:{u}+\mathrm{2}{i}\right)^{\mathrm{2}} −\left(\mathrm{3}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=\frac{−{i}}{\mathrm{4}}\int\frac{{du}}{\left(\sqrt{\mathrm{3}}\mathrm{sin}\:{u}+\mathrm{2}{i}−\mathrm{3}\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}\mathrm{sin}\:{u}+\mathrm{2}{k}+\mathrm{3}\sqrt{\mathrm{3}}\right)} \\ $$$$=\frac{−{i}}{\mathrm{4}\centerdot\mathrm{6}\sqrt{\mathrm{3}}}\left[\int\frac{{du}}{\left(\sqrt{\mathrm{3}}\mathrm{sin}\:{u}+\mathrm{2}{i}−\mathrm{3}\sqrt{\mathrm{3}}\right)}−\int\frac{{du}}{\sqrt{\mathrm{3}}\mathrm{sin}\:{u}+\mathrm{2}{i}+\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$$$=\frac{−{i}}{\mathrm{4}\centerdot\mathrm{6}\centerdot\mathrm{3}}\left[\int\frac{{du}}{\left(\mathrm{sin}\:{u}+\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{i}−\mathrm{3}\right)\right.}−\int\frac{{du}}{\mathrm{sin}\:{u}+\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{i}+\mathrm{3}\right)}\right] \\ $$

Answered by prakash jain last updated on 16/Nov/15

As given in comments both integrals are of the form ∫(du/(sin u+a)) This can be integrated using the substitution tan (u/2)=v du=((2dv)/((1+v^2 ))) sin u=((2v)/(1+v^2 )) ∫(du/(sin x+a))=∫ (((2dv)/(1+v^2 ))/(((2v)/(1+v^2 ))+a))=(1/a)∫ ((2dv)/(v^2 +2(v/a)+1)) =(1/a)∫ ((2dv)/((v+(1/a))^2 +((√((√a)−(1/a^2 ))))^2 ))=tan^(−1) ...(standard integral) Fill in the required value and do a reverse substitution. You will get a long result in tan^(−1) . You can also convert the result to ln or tanh^(−1) .

$$\mathrm{As}\:\mathrm{given}\:\mathrm{in}\:\mathrm{comments}\:\mathrm{both}\:\mathrm{integrals}\:\mathrm{are} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$$\int\frac{{du}}{\mathrm{sin}\:{u}+{a}} \\ $$$$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{integrated}\:\mathrm{using}\:\mathrm{the}\:\mathrm{substitution} \\ $$$$\mathrm{tan}\:\frac{{u}}{\mathrm{2}}={v} \\ $$$${du}=\frac{\mathrm{2}{dv}}{\left(\mathrm{1}+{v}^{\mathrm{2}} \right)} \\ $$$$\mathrm{sin}\:{u}=\frac{\mathrm{2}{v}}{\mathrm{1}+{v}^{\mathrm{2}} } \\ $$$$\int\frac{{du}}{\mathrm{sin}\:{x}+{a}}=\int\:\:\frac{\frac{\mathrm{2}{dv}}{\mathrm{1}+{v}^{\mathrm{2}} }}{\frac{\mathrm{2}{v}}{\mathrm{1}+{v}^{\mathrm{2}} }+{a}}=\frac{\mathrm{1}}{{a}}\int\:\frac{\mathrm{2}{dv}}{{v}^{\mathrm{2}} +\mathrm{2}\frac{{v}}{{a}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{a}}\int\:\frac{\mathrm{2}{dv}}{\left({v}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} +\left(\sqrt{\sqrt{{a}}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}\right)^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} ...\left(\mathrm{standard}\:\mathrm{integral}\right) \\ $$$$\mathrm{Fill}\:\mathrm{in}\:\mathrm{the}\:\mathrm{required}\:\mathrm{value}\:\mathrm{and}\:\mathrm{do}\:\mathrm{a}\:\mathrm{reverse} \\ $$$$\mathrm{substitution}.\:\mathrm{You}\:\mathrm{will}\:\mathrm{get}\:\mathrm{a}\:\mathrm{long}\:\mathrm{result}\:\mathrm{in}\:\mathrm{tan}^{−\mathrm{1}} . \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{also}\:\mathrm{convert}\:\mathrm{the}\:\mathrm{result}\:\mathrm{to}\:\mathrm{ln}\:\mathrm{or}\:\mathrm{tanh}^{−\mathrm{1}} . \\ $$

Commented by prakash jain last updated on 16/Nov/15

The result will contain i=(√(−1)) as well terms involving square roots of complex number.

$$\mathrm{The}\:\mathrm{result}\:\mathrm{will}\:\mathrm{contain}\:{i}=\sqrt{−\mathrm{1}}\:\mathrm{as}\:\mathrm{well}\:\mathrm{terms}\: \\ $$$$\mathrm{involving}\:\mathrm{square}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{number}. \\ $$

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