Đáp án:  `(sqrt[e^(π)]   +1)/2-ln((sqrt[e^(π)]+1)/(2))`

Giải thích các bước giải:

`\int_{0}^{π/2} (e^(2x).sinx+e^x.sinx-e^x)/(e^x+1) dx`

`=\int_{0}^{π/2} ([e^(x)]^2.sinx+e^x.sinx-e^x)/(e^x+1) dx`

`=\int_{0}^{π/2} (e^(x).sinx(e^x+1)-e^x)/(e^x+1) dx`

`=\int_{0}^{π/2} (e^(x).sinx(e^x+1))/(e^x+1) dx -\int_{0}^{π/2} (e^x)/(e^x+1) dx`

`=\int_{0}^{π/2} e^(x).sinx dx -\int_{0}^{π/2} (e^x)/(e^x+1) dx`

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+) Xét `I=\int_{0}^{π/2} e^(x).sinx dx`


Đặt: `{(sinx = u),(e^(x) dx = dv):}<=>{(cosxdx = du),(e^(x) = v):}`


`=> I=\int_{0}^{π/2} e^(x).sinx dx = e^(x) . sinx` $\bigg|^{π/2}_{0}$ `- \int_{0}^{π/2}e^(x).cosx dx`


Ta xét tiếp: `\int_{0}^{π/2}e^(x).cosx dx:`


Đặt: `{(cosx = a),(e^(x) dx = db):}<=>{(-sinxdx = da),(e^(x) = b):}`


`=>\int_{0}^{π/2}e^(x).cosx dx=e^(x).cosx` $\bigg|^{π/2}_{0}$ `-\int_{0}^{π/2}e^(x).(-sinx)dx `


`=e^(x).cosx`$\bigg|^{π/2}_{0}$`+\int_{0}^{π/2}e^(x).(sinx)dx=e^(x).cosx`$\bigg|^{π/2}_{0}$`+I `


Vậy `\int_{0}^{π/2}e^(x).cosx dx=e^(x).cosx`$\bigg|^{π/2}_{0}$`+I`


`=> I=\int_{0}^{π/2} e^(x).sinx dx = e^(x) . sinx`$\bigg|^{π/2}_{0}$` - \int_{0}^{π/2}e^(x).cosx dx`


`<=> I = e^(x) . sinx`$\bigg|^{π/2}_{0}$` - (e^(x).cosx`$\bigg|^{π/2}_{0}$`+I) `


`<=> I = e^(x) . sinx`$\bigg|^{π/2}_{0}$` - e^(x).cosx`$\bigg|^{π/2}_{0}$`-I `


`<=> 2I = e^(x) . sinx`$\bigg|^{π/2}_{0}$` - e^(x).cosx`$\bigg|^{π/2}_{0}$` `

`<=> 2I = e^(π/2) . sin(π/2) - e^(0) . sin0 - (e^(π/2).cos(π/2)-e^(0).cos0)`

`<=> 2I = [e^(π)]^(1/2) . 1 - 1.0 - ([e^(π)]^(1/2).0-1.1)`

`<=> 2I = sqrt[e^(π)]  - 0 +1`


`<=> I = (sqrt[e^(π)]   +1)/2 `


Vậy `\int_{0}^{π/2} e^(x).sinx dx= (sqrt[e^(π)]   +1)/2`

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+) Xét `\int_{0}^{π/2} (e^x)/(e^x+1) dx`

Đặt `e^x+1 = t`

`<=> e^xdx = dt`

=> `\int_{}^{} (e^x)/(e^x+1) dx=\int_{}^{} 1/t dt=ln|t|+c=ln|e^x+1|+C`

Vậy `\int_{0}^{π/2} (e^x)/(e^x+1) dx=ln|e^x+1|` $\bigg|^{π/2}_{0}$

`=ln|e^(π/2)+1|-ln|e^(0)+1|=ln|[e^(π)]^(1/2)+1|-ln|1+1|`

`=ln|sqrt[e^(π)]+1|-ln2=ln((sqrt[e^(π)]+1)/(2))` 

Vậy `\int_{0}^{π/2} (e^x)/(e^x+1) dx=ln((sqrt[e^(π)]+1)/(2))`

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Chứng minh ở trên ta có: `\int_{0}^{π/2} (e^(2x).sinx+e^x.sinx-e^x)/(e^x+1) dx`

`=\int_{0}^{π/2} e^(x).sinx dx -\int_{0}^{π/2} (e^x)/(e^x+1) dx`

`=(sqrt[e^(π)]   +1)/2-ln((sqrt[e^(π)]+1)/(2))`