Question Number 144764 by nonh1 last updated on 29/Jun/21
Answered by liberty last updated on 29/Jun/21
![∫_0 ^( (√(ln 2))) ∫_0 ^( 1) (1/2)xe^x^2 (((d(1+y^2 ))/(1+y^2 )))dx = ∫_0 ^(√(ln 2)) (1/2)xe^x^2 (ln (1+y^2 )∣_0 ^1 )dx =(1/4)ln 2∫_0 ^(√(ln 2)) e^x^2 d(x^2 ) = (1/4)ln 2 [ e^x^2 ]_0 ^(√(ln 2)) = (1/4)ln 2 [ e^(ln 2) −1 ] = (1/4)ln 2 [ 2−1 ]= (1/4)ln 2](https://www.tinkutara.com/question/Q144765.png)
$$\int_{\mathrm{0}} ^{\:\sqrt{\mathrm{ln}\:\mathrm{2}}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{xe}^{\mathrm{x}^{\mathrm{2}} } \:\left(\frac{\mathrm{d}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{ln}\:\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{xe}^{\mathrm{x}^{\mathrm{2}} } \left(\mathrm{ln}\:\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} \right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\mathrm{ln}\:\mathrm{2}}} \:\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\:\left[\:\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:\right]_{\mathrm{0}} ^{\sqrt{\mathrm{ln}\:\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\:\left[\:\mathrm{e}^{\mathrm{ln}\:\mathrm{2}} \:−\mathrm{1}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\:\left[\:\mathrm{2}−\mathrm{1}\:\right]=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2} \\ $$