, and 
![\int^1_0 \frac{2x+1}{e^x} dx=[(-2x-3)e^{-x}]^1_0](/tpl/images/0454/9267/08d7e.png)
![=[(-2(1)-3)e^{-(1)}]-[(-2(0)-3)e^{-(0)}]\\=[(-2-3)e^{-1}]-[(0-3)(1)]\\=-5e^{-1}+3\\=\frac{-5}{e}+3](/tpl/images/0454/9267/db38c.png)
Use integration by parts to find the integrals in exercise.
∫^1_0 2x+1/e^x dx....
Mathematics, 14.01.2020 21:31, 1060975
Use integration by parts to find the integrals in exercise.
∫^1_0 2x+1/e^x dx.
Answers: 3
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