Question Number 58250 by Tawa1 last updated on 20/Apr/19
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\:\left(\frac{\mathrm{3x}^{\mathrm{3}} \:−\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:−\:\mathrm{4}}{\sqrt{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:+\:\mathrm{2}}}\right)\:\mathrm{dx} \\ $$
Answered by MJS last updated on 21/Apr/19
![∫_0 ^1 ((3x^3 −x^2 +2x−4)/(√(x^2 −3x+2)))dx= [t=(√(x^2 −3x+2)) → dx=((−2tdt)/(√(4t^2 +1))); x=((3−(√(4t^2 +1)))/2)] =∫_(√2) ^0 (3t^2 +20−((25t^2 )/(√(4t^2 +1)))−((20)/(√(4t^2 +1))))dt= =[t^3 +20t−((25)/(16))(2t(√(4t^2 +1))−ln (2t+(√(4t^2 +1))))−10ln (2t+(√(4t^2 +1)))]_(√2) ^0 = =−[t^3 +20t−((25)/8)t(√(4t^2 +1))−((135)/(16))ln (2t+(√(4t^2 +1)))]_0 ^(√2) = =((135)/(16))ln (3+2(√2)) −((101(√2))/8)≈−2.98127](Q58295.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}\:\rightarrow\:{dx}=\frac{−\mathrm{2}{tdt}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}};\:{x}=\frac{\mathrm{3}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}\right] \\ $$$$=\underset{\sqrt{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{20}−\frac{\mathrm{25}{t}^{\mathrm{2}} }{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{20}}{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}\right){dt}= \\ $$$$=\left[{t}^{\mathrm{3}} +\mathrm{20}{t}−\frac{\mathrm{25}}{\mathrm{16}}\left(\mathrm{2}{t}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\right)\right)−\mathrm{10ln}\:\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\right)\right]_{\sqrt{\mathrm{2}}} ^{\mathrm{0}} = \\ $$$$=−\left[{t}^{\mathrm{3}} +\mathrm{20}{t}−\frac{\mathrm{25}}{\mathrm{8}}{t}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{135}}{\mathrm{16}}\mathrm{ln}\:\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} = \\ $$$$=\frac{\mathrm{135}}{\mathrm{16}}\mathrm{ln}\:\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{101}\sqrt{\mathrm{2}}}{\mathrm{8}}\approx−\mathrm{2}.\mathrm{98127} \\ $$