What is the integral of 1/(1.02^x) ?

The integral of 1/(1.02^x) is -50.49834…*1.02^{-x}+C

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integral of 1/(1.02^x)

\int \frac{1}{1.0 2 ^{x}} d x

- 50.49834 \dots \cdot 1.0 2^{- x} + C

Solution steps

\int \frac{1}{1.0 2 ^{x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{1.0 2 ^{x}} = 1.0 2^{- x}

= \int 1.0 2^{- x} d x

Apply u-substitution

= \int - 1.0 2^{u} d u

Take the constant out:

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

= - \int 1.0 2^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - \frac{1.0 2 ^{u}}{ln ( 1.02 )}

Substitute back

u = - x

= - \frac{1.0 2 ^{- x}}{ln ( 1.02 )}

Simplify

- \frac{1.0 2 ^{- x}}{ln ( 1.02 )} : - 50.49834 \dots \cdot 1.0 2^{- x}

= - 50.49834 \dots \cdot 1.0 2^{- x}

Add a constant to the solution

= - 50.49834 \dots \cdot 1.0 2^{- x} + C

What is the integral of 1/(1.02^x) ?
The integral of 1/(1.02^x) is -50.49834…*1.02^{-x}+C