Question Number 151010 by tabata last updated on 17/Aug/21
Commented by tabata last updated on 17/Aug/21
$${how}\:{can}\:{it}\:{solve}\:{this}\:? \\ $$
Answered by Olaf_Thorendsen last updated on 17/Aug/21
![I = ∫^1 _0 (x^3 /((x−1)^3 +3x−5)) dx I = ∫^1 _0 (x^3 /(x^3 −3x^2 +6x−6)) dx I = ∫^1 _0 (((x^3 −3x^2 +6x−6)+(3x^2 −6x+6))/(x^3 −3x^2 +6x−6)) dx I = ∫^1 _0 1+(((3x^2 −6x+6)/(x^3 −3x^2 +6x−6))) dx I = [x+ln∣x^3 −3x^2 +6x−6∣]_0 ^1 I = 1+ln2−ln6 I = 1−ln3](https://www.tinkutara.com/question/Q151016.png)
$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{5}}\:{dx} \\ $$$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{6}}\:{dx} \\ $$$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{6}\right)+\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{6}}\:{dx} \\ $$$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \mathrm{1}+\left(\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{6}}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:\left[{x}+\mathrm{ln}\mid{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{6}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\mathrm{I}\:=\:\mathrm{1}+\mathrm{ln2}−\mathrm{ln6} \\ $$$$\mathrm{I}\:=\:\mathrm{1}−\mathrm{ln3} \\ $$
Commented by peter frank last updated on 18/Aug/21
$${thank}\:{you} \\ $$