What is the integral of 1/3 e^{-x/2}(x+1) ?

The integral of 1/3 e^{-x/2}(x+1) is 1/3 (-2e^{-x/2}x-6e^{-x/2})+C

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integral of 1/3 e^{-x/2}(x+1)

\int \frac{1}{3} e^{- \frac{x}{2}} (x + 1) d x

\frac{1}{3} (- 2 e^{- \frac{x}{2}} x - 6 e^{- \frac{x}{2}}) + C

Solution steps

\int \frac{1}{3} e^{- \frac{x}{2}} (x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int e^{- \frac{x}{2}} (x + 1) d x

Apply Integration By Parts

: - 2 e^{- \frac{x}{2}} (x + 1) - \int - 2 e^{- \frac{x}{2}} d x

= \frac{1}{3} (- 2 e^{- \frac{x}{2}} (x + 1) - \int - 2 e^{- \frac{x}{2}} d x)

\int - 2 e^{- \frac{x}{2}} d x = 4 e^{- \frac{x}{2}}

= \frac{1}{3} (- 2 e^{- \frac{x}{2}} (x + 1) - 4 e^{- \frac{x}{2}})

Simplify

\frac{1}{3} (- 2 e^{- \frac{x}{2}} (x + 1) - 4 e^{- \frac{x}{2}}) : \frac{1}{3} (- 2 e^{- \frac{x}{2}} x - 6 e^{- \frac{x}{2}})

= \frac{1}{3} (- 2 e^{- \frac{x}{2}} x - 6 e^{- \frac{x}{2}})

Add a constant to the solution

= \frac{1}{3} (- 2 e^{- \frac{x}{2}} x - 6 e^{- \frac{x}{2}}) + C

\int sec^{8} (t) an x d x

\int - \frac{1}{9} x^{- \frac{10}{9}} d x

\frac{d}{d x} (\frac{2}{x ^{13}})

\int (tan (x) - sin (x) sin (y)) d x

f (x) = sin (x), at x = 1

What is the integral of 1/3 e^{-x/2}(x+1) ?
The integral of 1/3 e^{-x/2}(x+1) is 1/3 (-2e^{-x/2}x-6e^{-x/2})+C