Question Number 175697 by Tawa11 last updated on 05/Sep/22
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{3x}^{\mathrm{3}} \:−\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:−\:\mathrm{4}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:+\:\mathrm{2}}}\:\mathrm{dx} \\ $$
Answered by MJS_new last updated on 05/Sep/22
![∫_0 ^1 (((x−1)(3x^2 +2x+4))/( (√((x−2)(x−1)))))dx= =−∫_0 ^1 ((√(x−1))/( (√(x−2))))(3x^2 +2x+4)dx −∫((√(x−1))/( (√(x−2))))(3x^2 +2x+4)dx= [t=((√(x−2))/( (√(x+1)))) → dx=2(√((x−2)(x−1)^3 ))dt] =−2∫((9t^4 −26t^2 +20)/((t^2 −1)^4 ))dt=... =((t(135t^4 −312t^2 +185))/(8(t^2 −1)^3 ))+((135)/(16))+ln ((t−1)/(t+1)) = =−(((√()x−2)(x−1)))(8x^2 +26x+101))/8)+((135)/(16))ln ∣2x−3−2(√((x−2)(x−1)))∣ +C ⇒ answer is −((101(√2)+135ln (−1+(√2)))/8)](https://www.tinkutara.com/question/Q175711.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}−\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{\:\sqrt{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}}{dx}= \\ $$$$=−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}−\mathrm{2}}}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right){dx} \\ $$$$ \\ $$$$−\int\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}−\mathrm{2}}}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right){dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{{x}−\mathrm{2}}}{\:\sqrt{{x}+\mathrm{1}}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=−\mathrm{2}\int\frac{\mathrm{9}{t}^{\mathrm{4}} −\mathrm{26}{t}^{\mathrm{2}} +\mathrm{20}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{dt}=… \\ $$$$=\frac{{t}\left(\mathrm{135}{t}^{\mathrm{4}} −\mathrm{312}{t}^{\mathrm{2}} +\mathrm{185}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{135}}{\mathrm{16}}+\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:= \\ $$$$=−\frac{\sqrt{\left.\right)\left.{x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}\left(\mathrm{8}{x}^{\mathrm{2}} +\mathrm{26}{x}+\mathrm{101}\right)}{\mathrm{8}}+\frac{\mathrm{135}}{\mathrm{16}}\mathrm{ln}\:\mid\mathrm{2}{x}−\mathrm{3}−\mathrm{2}\sqrt{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}\mid\:+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:−\frac{\mathrm{101}\sqrt{\mathrm{2}}+\mathrm{135ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}} \\ $$
Commented by Tawa11 last updated on 05/Sep/22
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Answered by Frix last updated on 05/Sep/22
![∫_0 ^1 ((3x^3 −x^2 +2x−4)/( (√(x^2 −3x+2))))dx=_(dx=(dt/2)) ^(t=2x−3) =(1/8)∫_(−3) ^(−1) ((3t^3 +25t^2 +77t+55)/( (√(t^2 −1))))dt=_(dt=((√(t^2 −1))/u)dt) ^(u=t+(√(t^2 −1))) =∫_(−3+2(√2)) ^(−1) (((3u^2 )/(64))+((25u)/(32))+((317)/(64))+((135)/(16u))+((317)/(64u^2 ))+((25)/(32u^3 ))+(3/(64u^4 )))du= =[(u^3 /(64))+((25u^2 )/(64))+((317u)/(64))+((135ln ∣u∣)/(16))−((317)/(64u))−((25)/(64u^2 ))−(1/(64u^3 ))]_(−3+2(√2)) ^(−1) = =−((101(√2)+135ln (−1+(√2)))/8)](https://www.tinkutara.com/question/Q175732.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}\underset{{dx}=\frac{{dt}}{\mathrm{2}}} {\overset{{t}=\mathrm{2}{x}−\mathrm{3}} {=}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{−\mathrm{3}} {\overset{−\mathrm{1}} {\int}}\frac{\mathrm{3}{t}^{\mathrm{3}} +\mathrm{25}{t}^{\mathrm{2}} +\mathrm{77}{t}+\mathrm{55}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{dt}\underset{{dt}=\frac{\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{{u}}{dt}} {\overset{{u}={t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} {=}} \\ $$$$=\underset{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} {\overset{−\mathrm{1}} {\int}}\left(\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{64}}+\frac{\mathrm{25}{u}}{\mathrm{32}}+\frac{\mathrm{317}}{\mathrm{64}}+\frac{\mathrm{135}}{\mathrm{16}{u}}+\frac{\mathrm{317}}{\mathrm{64}{u}^{\mathrm{2}} }+\frac{\mathrm{25}}{\mathrm{32}{u}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{64}{u}^{\mathrm{4}} }\right){du}= \\ $$$$=\left[\frac{{u}^{\mathrm{3}} }{\mathrm{64}}+\frac{\mathrm{25}{u}^{\mathrm{2}} }{\mathrm{64}}+\frac{\mathrm{317}{u}}{\mathrm{64}}+\frac{\mathrm{135ln}\:\mid{u}\mid}{\mathrm{16}}−\frac{\mathrm{317}}{\mathrm{64}{u}}−\frac{\mathrm{25}}{\mathrm{64}{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{64}{u}^{\mathrm{3}} }\right]_{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} ^{−\mathrm{1}} = \\ $$$$=−\frac{\mathrm{101}\sqrt{\mathrm{2}}+\mathrm{135ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
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