What is the integral of (1.8^{0.045x}-1) ?

The integral of (1.8^{0.045x}-1) is 37.80661…*1.02680…^x-x+C

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integral of (1.8^{0.045x}-1)

\int (1. 8^{0.045 x} - 1) d x

37.80661 \dots \cdot 1.02680 \dots^{x} - x + C

Solution steps

\int 1. 8^{0.045 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

= \int 1. 8^{0.045 x} d x - \int 1 d x

\int 1. 8^{0.045 x} d x = 37.80661 \dots \cdot 1.02680 \dots^{x}

\int 1 d x = x

= 37.80661 \dots \cdot 1.02680 \dots^{x} - x

Add a constant to the solution

= 37.80661 \dots \cdot 1.02680 \dots^{x} - x + C

f (x) = e^{\frac{x}{3}}

t^{2} y^{^{''}} + 4 t y^{^{'}} + 2 y = 0

x \to 0 + lim ((x^{2}) ln (x))

\int \frac{2 ^{x}}{2 ^{x} + 1} d x

What is the integral of (1.8^{0.045x}-1) ?
The integral of (1.8^{0.045x}-1) is 37.80661…*1.02680…^x-x+C