What is the integral of (5e^{-x}-3/2)^2 ?

The integral of (5e^{-x}-3/2)^2 is -25/2 e^{-2x}+15e^{-x}+9/4 x+C

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integral of (5e^{-x}-3/2)^2

\int (5 e^{- x} - \frac{3}{2})^{2} d x

- \frac{25}{2} e^{- 2 x} + 15 e^{- x} + \frac{9}{4} x + C

Solution steps

\int (5 e^{- x} - \frac{3}{2})^{2} d x

Simplify

(5 e^{- x} - \frac{3}{2})^{2} : 25 e^{- 2 x} - 15 e^{- x} + \frac{9}{4}

= \int 25 e^{- 2 x} - 15 e^{- x} + \frac{9}{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 25 e^{- 2 x} d x - \int 15 e^{- x} d x + \int \frac{9}{4} d x

\int 25 e^{- 2 x} d x = - \frac{25}{2} e^{- 2 x}

\int 15 e^{- x} d x = - 15 e^{- x}

\int \frac{9}{4} d x = \frac{9}{4} x

= - \frac{25}{2} e^{- 2 x} - (- 15 e^{- x}) + \frac{9}{4} x

Simplify

= - \frac{25}{2} e^{- 2 x} + 15 e^{- x} + \frac{9}{4} x

Add a constant to the solution

= - \frac{25}{2} e^{- 2 x} + 15 e^{- x} + \frac{9}{4} x + C

\frac{d ^{2}}{d x ^{2}} (ln (1 + x))

f (x) = 6 e^{x} + x, at x = 0

f (x) = 2 x x, at x = 1

y = 2 x, y = 2 x

y^{^{''}} - y^{^{'}} - 2 y = e^{x} + x

What is the integral of (5e^{-x}-3/2)^2 ?
The integral of (5e^{-x}-3/2)^2 is -25/2 e^{-2x}+15e^{-x}+9/4 x+C