Evaluate the integral: integral (2^(x+1)
Evaluate the integral: $$ \int \frac{2^{x+1} - 5^{x-1}}{10^x} \, dx. $$
Explanation of Method
Shifted exponents like $2^{x+1}$ split into a constant factor times $2^x$ via $a^{b+c}=a^b a^c$. Dividing each resulting term by $10^x$ and using $a^m/a^n = a^{m-n}$ collapses each fraction into a single exponential with a new (often fractional) base, ready for the general antiderivative formula.
- Split each shifted exponent into a constant factor times a pure power.
- Divide each term by the common denominator base, combining into new bases.
- Integrate each resulting exponential term.
- Combine the results and add the constant of integration.
Step 1 · Simplify the terms
The integrand is:
$$ \frac{2^{x+1} - 5^{x-1}}{10^x}. $$
Using the property of exponents $a^{b+c} = a^b\cdot a^c$, rewrite $2^{x+1}$ and $5^{x-1}$:
$$ 2^{x+1} = 2^x\cdot 2^1 = 2\cdot 2^x, \qquad 5^{x-1} = 5^x\cdot 5^{-1} = \frac{5^x}{5}. $$
Substitute these into the integrand:
$$ \frac{2^{x+1}-5^{x-1}}{10^x} = \frac{2\cdot 2^x}{10^x} - \frac{\frac{5^x}{5}}{10^x}. $$
Using the property of exponents $\dfrac{a^m}{a^n} = a^{m-n}$, simplify each term:
$$ \frac{2\cdot 2^x}{10^x} = 2\cdot \frac{2^x}{10^x} = 2\cdot \left(\frac{2}{10}\right)^x = 2\cdot \left(\frac{1}{5}\right)^x, $$
$$ \frac{\frac{5^x}{5}}{10^x} = \frac{5^x}{5\cdot 10^x} = \frac{1}{5}\cdot \frac{5^x}{10^x} = \frac{1}{5}\cdot \left(\frac{5}{10}\right)^x = \frac{1}{5}\cdot \left(\frac{1}{2}\right)^x. $$
Thus, the integrand becomes:
$$ \frac{2^{x+1}-5^{x-1}}{10^x} = 2\cdot \left(\frac{1}{5}\right)^x - \frac{1}{5}\cdot \left(\frac{1}{2}\right)^x. $$
Step 2 · Rewrite the integral
The integral can now be written as:
$$ \int \frac{2^{x+1}-5^{x-1}}{10^x}\,dx = \int 2\cdot \left(\frac{1}{5}\right)^x dx - \int \frac{1}{5}\cdot \left(\frac{1}{2}\right)^x dx. $$
Step 3 · Solve each term
We evaluate each integral separately using the formula for the integral of an exponential function:
$$ \int a^x\,dx = \frac{a^x}{\ln(a)} + C, \quad \text{for } a>0,\ a\neq 1. $$
First term: $\displaystyle \int 2\cdot \left(\frac{1}{5}\right)^x dx$. Let $a = \dfrac{1}{5}$. Using the constant multiple rule:
$$ \int 2\cdot \left(\frac{1}{5}\right)^x dx = 2\cdot \int \left(\frac{1}{5}\right)^x dx = 2\cdot \frac{\left(\frac{1}{5}\right)^x}{\ln\left(\frac{1}{5}\right)}. $$
Simplify $\ln\left(\dfrac{1}{5}\right) = -\ln(5)$:
$$ \int 2\cdot \left(\frac{1}{5}\right)^x dx = -\frac{2\cdot \left(\frac{1}{5}\right)^x}{\ln(5)}. $$
Second term: $\displaystyle \int \frac{1}{5}\cdot \left(\frac{1}{2}\right)^x dx$. Let $a = \dfrac{1}{2}$. Using the constant multiple rule:
$$ \int \frac{1}{5}\cdot \left(\frac{1}{2}\right)^x dx = \frac{1}{5}\cdot \int \left(\frac{1}{2}\right)^x dx = \frac{1}{5}\cdot \frac{\left(\frac{1}{2}\right)^x}{\ln\left(\frac{1}{2}\right)}. $$
Simplify $\ln\left(\dfrac{1}{2}\right) = -\ln(2)$:
$$ \int \frac{1}{5}\cdot \left(\frac{1}{2}\right)^x dx = -\frac{\left(\frac{1}{2}\right)^x}{5\cdot \ln(2)}. $$
Step 4 · Combine the results
Adding the two results, the solution to the integral is:
$$ \int \frac{2^{x+1}-5^{x-1}}{10^x}\,dx = -\frac{2\cdot \left(\frac{1}{5}\right)^x}{\ln(5)} - \frac{\left(\frac{1}{2}\right)^x}{5\cdot \ln(2)} + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Rewritten as | Antiderivative |
|---|---|---|
| $2^{x+1}/10^x$ | $2\cdot(1/5)^x$ | $-\dfrac{2\cdot(1/5)^x}{\ln 5}$ |
| $5^{x-1}/10^x$ | $\tfrac{1}{5}\cdot(1/2)^x$ | $-\dfrac{(1/2)^x}{5\ln 2}$ (subtracted) |
Additional Notes
- The integrand is defined for all real $x$, since exponential functions with positive base are defined everywhere.
- Because $\ln(1/5)=-\ln 5$ and $\ln(1/2)=-\ln 2$ are both negative, both antiderivative terms pick up an extra minus sign, which is why the final answer has two negative terms even though one term in the integrand was subtracted.
- The result can be checked by differentiating $-\dfrac{2\cdot(1/5)^x}{\ln 5} - \dfrac{(1/2)^x}{5\ln 2}$ and confirming it returns the original integrand.
Result
The solution to the integral is: