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Find the integral value
Answer
$- \dfrac {x + 1} {e ^ {x}} + C$
Using partial integration
$\color{#FF6800}{\int xe ^ {- x}dx}$
$ $ To solve the integral, substitute $ v = - e ^ {- x} $ and $ u = x $ , and use the partial integration formula $ \int udv = uv - \int vdu $ $ $
$\color{#FF6800}{x \times (- e ^ {- x}) - \int - e ^ {- x}dx}$
$\color{#FF6800}{}x \times (- e ^ {- x})\color{#FF6800}{ - }\int \color{#FF6800}{- }e ^ {- x}\color{#FF6800}{}dx\color{#FF6800}{}$
$ $ Utilize integration $ \int - f(x)dx = - \int f(x)dx $ $ $
$\color{#FF6800}{}x \times (- e ^ {- x})\color{#FF6800}{ + }\int e ^ {- x}dx\color{#FF6800}{}$
$\color{#FF6800}{}x \times (- e ^ {- x})\color{#FF6800}{ + \int e ^ {- x}dx}$
$ $ Using $ \int e ^ {- x}dx = - e ^ {- x} $ , calculate the integral $ $
$\color{#FF6800}{}x \times (- e ^ {- x})\color{#FF6800}{ - e ^ {- x}}$
$\color{#FF6800}{x \times (- e ^ {- x}) - e ^ {- x}}$
$ $ Solve the formula $ $
$\color{#FF6800}{- \dfrac {x + 1} {e ^ {x}}}$
$\color{#FF6800}{- \dfrac {x + 1} {e ^ {x}}}$
$ $ Add integral constant $ C \in ℝ $ $ $
$\color{#FF6800}{- \dfrac {x + 1} {e ^ {x}} + C}$
![search-thumbnail-18. The solution of the equation 2x+3 (x)-4 =4(whe (where [x] and (\times3 denote integral and](https://thumb-m.mathpresso.io/qanda-thumbnail-storage/questions/Z52UJyf1wSr1U7rCMzC4pieKhd4hlxtwVcA6CrF2bIRrfzVCm3.jpg?target_format=webp&width=652)
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