What is the integral of ((3x(e^{-2x}-7x))/5) ?

The integral of ((3x(e^{-2x}-7x))/5) is -3/10 e^{-2x}x-3/20 e^{-2x}-(7x^3)/5+C

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integral of ((3x(e^{-2x}-7x))/5)

\int (\frac{3 x ( e ^{- 2 x} - 7 x )}{5}) d x

- \frac{3}{10} e^{- 2 x} x - \frac{3}{20} e^{- 2 x} - \frac{7 x ^{3}}{5} + C

Solution steps

\int \frac{3 x ( e ^{- 2 x} - 7 x )}{5} d x

Take the constant out:

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

= \frac{3}{5} \cdot \int x (e^{- 2 x} - 7 x) d x

Expand

x (e^{- 2 x} - 7 x) : e^{- 2 x} x - 7 x^{2}

= \frac{3}{5} \cdot \int e^{- 2 x} x - 7 x^{2} 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

= \frac{3}{5} (\int e^{- 2 x} x d x - \int 7 x^{2} d x)

\int e^{- 2 x} x d x = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x})

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

= \frac{3}{5} (\frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) - \frac{7 x ^{3}}{3})

Simplify

\frac{3}{5} (\frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) - \frac{7 x ^{3}}{3}) : - \frac{3}{10} e^{- 2 x} x - \frac{3}{20} e^{- 2 x} - \frac{7 x ^{3}}{5}

= - \frac{3}{10} e^{- 2 x} x - \frac{3}{20} e^{- 2 x} - \frac{7 x ^{3}}{5}

Add a constant to the solution

= - \frac{3}{10} e^{- 2 x} x - \frac{3}{20} e^{- 2 x} - \frac{7 x ^{3}}{5} + C

x \to \infty lim ((\frac{1}{x})^{\frac{1}{x} + 1})

\int_{2}^{e^{3}} \frac{1}{x} d x

y = \frac{2 cos ( x )}{x + 1}

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

e^{- 3 x^{2}}

What is the integral of ((3x(e^{-2x}-7x))/5) ?
The integral of ((3x(e^{-2x}-7x))/5) is -3/10 e^{-2x}x-3/20 e^{-2x}-(7x^3)/5+C