What is the integral of (10x+1)e^{-5x} ?

The integral of (10x+1)e^{-5x} is -2e^{-5x}x-3/5 e^{-5x}+C

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integral of (10x+1)e^{-5x}

\int (10 x + 1) e^{- 5 x} d x

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

Solution steps

\int (10 x + 1) e^{- 5 x} d x

Expand

(10 x + 1) e^{- 5 x} : 10 e^{- 5 x} x + e^{- 5 x}

= \int 10 e^{- 5 x} x + e^{- 5 x} 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 10 e^{- 5 x} x d x + \int e^{- 5 x} d x

\int 10 e^{- 5 x} x d x = - 2 e^{- 5 x} x - \frac{2}{5} e^{- 5 x}

\int e^{- 5 x} d x = - \frac{1}{5} e^{- 5 x}

= - 2 e^{- 5 x} x - \frac{2}{5} e^{- 5 x} - \frac{1}{5} e^{- 5 x}

Add similar elements:

- \frac{2}{5} e^{- 5 x} - \frac{1}{5} e^{- 5 x} = - \frac{3}{5} e^{- 5 x}

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

Add a constant to the solution

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

\int cos (3 x - 2 π) cos (x + π) d x

2 (x^{3} - x)^{4}

n = 2 \sum \infty \frac{2}{n ^{3}}

n = 0 \sum \infty 2^{2 n}

\frac{d}{d x} (a^{f (x)})

What is the integral of (10x+1)e^{-5x} ?
The integral of (10x+1)e^{-5x} is -2e^{-5x}x-3/5 e^{-5x}+C