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Question Number 151010 by tabata last updated on 17/Aug/21

Commented by tabata last updated on 17/Aug/21

$$howcanitsolvethis?$$

Answered by Olaf_Thorendsen last updated on 17/Aug/21

$$\mathrm{I}={\underset{0}{\int }}^1\frac{x^3}{{(x-1)}^3+3x-5}dx$$$$\mathrm{I}={\underset{0}{\int }}^1\frac{x^3}{x^3-3x^2+6x-6}dx$$$$\mathrm{I}={\underset{0}{\int }}^1\frac{(x^3-3x^2+6x-6)+(3x^2-6x+6)}{x^3-3x^2+6x-6}dx$$$$\mathrm{I}={\underset{0}{\int }}^{1}1+\left(\frac{3x^2-6x+6}{x^3-3x^2+6x-6}\right)dx$$$$\mathrm{I}={[x+\ln \mid x^3-3x^2+6x-6\mid ]}_0^1$$$$\mathrm{I}=1+\ln 2-\ln 6$$$$\mathrm{I}=1-\ln 3$$

Commented by peter frank last updated on 18/Aug/21

$$thankyou$$

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