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Question Number 11864 by ankhaaankhaa last updated on 03/Apr/17

∫_7 ^(10) x^2 dx/x^2 −3x+2=

$$\underset{\mathrm{7}} {\overset{\mathrm{10}} {\int}}{x}^{\mathrm{2}} {dx}/{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}= \\ $$

Answered by ajfour last updated on 03/Apr/17

∫(x^2 /(x^2 −3x+2))dx = ∫ ((x^2 −3x+2+(3x−2))/(x^2 −3x+2))dx = ∫dx + ∫((3x−2)/((x−1)(x−2))) dx = x+ 4∫ (dx/(x−2)) −∫ (dx/(x−1)) +c = x+ 4ln ∣x−2∣−ln ∣x−1∣+ C ∫_7 ^( 10) (x^2 /(x^2 −3x+2))dx = (7−3)+4ln (((10−2)/(7−2)))−ln (((10−1)/(7−1))) = 4+4ln (8/5)−ln (3/2) .

$$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{dx}\:=\:\int\:\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}+\left(\mathrm{3}{x}−\mathrm{2}\right)}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{dx} \\ $$$$=\:\int{dx}\:+\:\int\frac{\mathrm{3}{x}−\mathrm{2}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}\:{dx} \\ $$$$=\:{x}+\:\mathrm{4}\int\:\frac{{dx}}{{x}−\mathrm{2}}\:−\int\:\frac{{dx}}{{x}−\mathrm{1}}\:+{c} \\ $$$$\:\:=\:{x}+\:\mathrm{4ln}\:\mid{x}−\mathrm{2}\mid−\mathrm{ln}\:\mid{x}−\mathrm{1}\mid+\:{C} \\ $$$$\int_{\mathrm{7}} ^{\:\:\mathrm{10}} \frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{dx} \\ $$$$\:\:\:\:=\:\left(\mathrm{7}−\mathrm{3}\right)+\mathrm{4ln}\:\left(\frac{\mathrm{10}−\mathrm{2}}{\mathrm{7}−\mathrm{2}}\right)−\mathrm{ln}\:\left(\frac{\mathrm{10}−\mathrm{1}}{\mathrm{7}−\mathrm{1}}\right) \\ $$$$=\:\mathrm{4}+\mathrm{4ln}\:\left(\mathrm{8}/\mathrm{5}\right)−\mathrm{ln}\:\left(\mathrm{3}/\mathrm{2}\right)\:. \\ $$

Answered by sma3l2996 last updated on 03/Apr/17

A=∫_7 ^(10) ((x^2 dx)/(x^2 −3x+2))=∫_7 ^(10) (1+((3x−2)/(x^2 −3x+2)))dx =[x]_7 ^(10) +(3/2)∫_7 ^(10) ((2x−(4/3))/(x^2 −3x+2))dx =3+(3/2)∫_7 ^(10) ((2x−3)/(x^2 −3x+2))dx+(5/2)∫_7 ^(10) (dx/((x−1)(x−2))) =3+(3/2)[ln∣x^2 −3x+2∣]_7 ^(10) +(5/2)∫_7 ^(10) ((1/(x−2))−(1/(x−1)))dx =3+(3/2)ln(((36)/(15)))+(5/2)[ln∣((x−2)/(x−1))∣]_7 ^(10) =(3/2)ln(((12)/5))+(5/2)(ln((8/9))−ln((5/6)))+3 A=(3/2)ln(((12)/5))+(5/2)ln(((16)/(15)))+3

$${A}=\int_{\mathrm{7}} ^{\mathrm{10}} \frac{{x}^{\mathrm{2}} {dx}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}=\int_{\mathrm{7}} ^{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{3}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}\right){dx} \\ $$$$=\left[{x}\right]_{\mathrm{7}} ^{\mathrm{10}} +\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{7}} ^{\mathrm{10}} \frac{\mathrm{2}{x}−\frac{\mathrm{4}}{\mathrm{3}}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{dx} \\ $$$$=\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{7}} ^{\mathrm{10}} \frac{\mathrm{2}{x}−\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{dx}+\frac{\mathrm{5}}{\mathrm{2}}\int_{\mathrm{7}} ^{\mathrm{10}} \frac{{dx}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)} \\ $$$$=\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}\left[{ln}\mid{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\mid\right]_{\mathrm{7}} ^{\mathrm{10}} +\frac{\mathrm{5}}{\mathrm{2}}\int_{\mathrm{7}} ^{\mathrm{10}} \left(\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{{x}−\mathrm{1}}\right){dx} \\ $$$$=\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\frac{\mathrm{36}}{\mathrm{15}}\right)+\frac{\mathrm{5}}{\mathrm{2}}\left[{ln}\mid\frac{{x}−\mathrm{2}}{{x}−\mathrm{1}}\mid\right]_{\mathrm{7}} ^{\mathrm{10}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\frac{\mathrm{12}}{\mathrm{5}}\right)+\frac{\mathrm{5}}{\mathrm{2}}\left({ln}\left(\frac{\mathrm{8}}{\mathrm{9}}\right)−{ln}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)+\mathrm{3} \\ $$$${A}=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\frac{\mathrm{12}}{\mathrm{5}}\right)+\frac{\mathrm{5}}{\mathrm{2}}{ln}\left(\frac{\mathrm{16}}{\mathrm{15}}\right)+\mathrm{3} \\ $$

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