Question Number 62192 by Sardor2211 last updated on 17/Jun/19
Answered by mr W last updated on 17/Jun/19
![∫_0 ^1 (x^3 /(x^2 +1))dx =(1/2)∫_0 ^1 (x^2 /(x^2 +1))dx^2 =(1/2)∫_0 ^1 [1−(1/(x^2 +1))]dx^2 =(1/2)[x^2 −ln (x^2 +1)]_0 ^1 =(1/2)[1−ln 2]](https://www.tinkutara.com/question/Q62193.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right]{dx}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}^{\mathrm{2}} −\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\mathrm{ln}\:\mathrm{2}\right] \\ $$