Question

$\int \frac{1+4^x+16^x}{1+2^x+4^x} dx$

Answer 1

$\frac{4^x}{2 \ln 2} - \frac{2^x}{\ln 2} + x + C$

Answer 2

The given integral is: $\int \frac{1+4^x+16^x}{1+2^x+4^x} dx$

Step 1: Simplify the integrand using substitution. Let $y = 2^x$. Then $4^x = (2^x)^2 = y^2$ and $16^x = (2^x)^4 = y^4$.

Substitute these into the integrand: $\frac{1+y^2+y^4}{1+y+y^2}$

Step 2: Apply algebraic identity to the numerator. Recall the algebraic identity for sum of powers: $a^4+a^2+1 = (a^2+1)^2 - a^2 = (a^2+1-a)(a^2+1+a)$. Applying this identity with $a=y$, the numerator becomes: $1+y^2+y^4 = (y^2-y+1)(y^2+y+1)$

Step 3: Simplify the rational expression. Substitute the factored numerator back into the integrand expression: $\frac{(y^2-y+1)(y^2+y+1)}{1+y+y^2}$ Since $1+y+y^2$ is common to both numerator and denominator, we can cancel it out (assuming $1+y+y^2 \neq 0$, which is always true for real $y=2^x > 0$). The expression simplifies to: $y^2-y+1$

Step 4: Substitute back $x$ terms. Substitute $y = 2^x$ back into the simplified expression: $(2^x)^2 - 2^x + 1 = 4^x - 2^x + 1$

Step 5: Integrate the simplified expression. Now, we need to integrate $4^x - 2^x + 1$ with respect to $x$: $\int (4^x - 2^x + 1) dx$ Using the standard integral formula $\int a^x dx = \frac{a^x}{\ln a} + C$: $\int 4^x dx = \frac{4^x}{\ln 4}$ $\int 2^x dx = \frac{2^x}{\ln 2}$ $\int 1 dx = x$ Combining these results: $\frac{4^x}{\ln 4} - \frac{2^x}{\ln 2} + x + C$

Step 6: Simplify the final result. We know that $\ln 4 = \ln (2^2) = 2 \ln 2$. Substitute this into the expression: $\frac{4^x}{2 \ln 2} - \frac{2^x}{\ln 2} + x + C$ To combine the terms with $\ln 2$ in the denominator: $\frac{4^x - 2 \cdot 2^x}{2 \ln 2} + x + C$ $\frac{4^x - 2^{x+1}}{2 \ln 2} + x + C$

The final answer is: $\frac{4^x}{2 \ln 2} - \frac{2^x}{\ln 2} + x + C$ or $\frac{4^x - 2^{x+1}}{2 \ln 2} + x + C$