What is the integral of 12(e^{0.02x}-1) ?

The integral of 12(e^{0.02x}-1) is 12(50e^{0.02x}-x)+C

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integral of 12(e^{0.02x}-1)

\int 12 (e^{0.02 x} - 1) d x

12 (50 e^{0.02 x} - x) + C

Solution steps

\int 12 (e^{0.02 x} - 1) d x

Take the constant out:

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

= 12 \cdot \int e^{0.02 x} - 1 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

= 12 (\int e^{0.02 x} d x - \int 1 d x)

\int e^{0.02 x} d x = 50 e^{0.02 x}

\int 1 d x = x

= 12 (50 e^{0.02 x} - x)

Add a constant to the solution

= 12 (50 e^{0.02 x} - x) + C

\int \frac{1}{9 e ^{- 2 u} - 1} d u

2 x^{- 5}

\int - 2 cos (x) d x

\int_{1}^{6} x - 1 d x

y = (13 x^{2} - 26 x + 26) e^{- 8 x}

What is the integral of 12(e^{0.02x}-1) ?
The integral of 12(e^{0.02x}-1) is 12(50e^{0.02x}-x)+C