Question

If $\int \frac{e^x}{e^{2x}+6e^x+5}dx = -\frac{1}{4} \log_e (\frac{e^x+a}{e^x+b})+c$, then $\frac{|a-b|}{4}$ is ...

• A. 1
• B. 3
• C. 4
• D. 15

Answer 1

Answer: (A)

1

Answer 2

To solve the integral $\int \frac{e^x}{e^{2x}+6e^x+5}dx$, we use a substitution.

Let $t = e^x$. Then, $dt = e^x dx$.

Substitute these into the integral: $\int \frac{e^x}{e^{2x}+6e^x+5}dx = \int \frac{dt}{t^2+6t+5}$

Now, factor the quadratic in the denominator: $t^2+6t+5 = (t+1)(t+5)$

So the integral becomes: $\int \frac{dt}{(t+1)(t+5)}$

We use partial fraction decomposition for the integrand $\frac{1}{(t+1)(t+5)}$. Let $\frac{1}{(t+1)(t+5)} = \frac{A}{t+1} + \frac{B}{t+5}$ Multiply both sides by $(t+1)(t+5)$: $1 = A(t+5) + B(t+1)$

To find $A$, set $t=-1$: $1 = A(-1+5) + B(-1+1)$ $1 = 4A \implies A = \frac{1}{4}$

To find $B$, set $t=-5$: $1 = A(-5+5) + B(-5+1)$ $1 = -4B \implies B = -\frac{1}{4}$

Substitute the values of $A$ and $B$ back into the partial fraction form: $\frac{1}{(t+1)(t+5)} = \frac{1/4}{t+1} - \frac{1/4}{t+5}$

Now, integrate: $\int \left( \frac{1/4}{t+1} - \frac{1/4}{t+5} \right) dt = \frac{1}{4} \int \left( \frac{1}{t+1} - \frac{1}{t+5} \right) dt$ $= \frac{1}{4} \left( \log_e |t+1| - \log_e |t+5| \right) + c$ Using the logarithm property $\log X - \log Y = \log(X/Y)$: $= \frac{1}{4} \log_e \left| \frac{t+1}{t+5} \right| + c$

Substitute back $t = e^x$. Since $e^x > 0$, $e^x+1$ and $e^x+5$ are always positive, so we can remove the absolute value signs: $\int \frac{e^x}{e^{2x}+6e^x+5}dx = \frac{1}{4} \log_e \left( \frac{e^x+1}{e^x+5} \right) + c$

The problem states that the integral is equal to $-\frac{1}{4} \log_e (\frac{e^x+a}{e^x+b})+c$. We need to equate our result with the given form: $\frac{1}{4} \log_e \left( \frac{e^x+1}{e^x+5} \right) = -\frac{1}{4} \log_e \left( \frac{e^x+a}{e^x+b} \right)$ Multiply both sides by 4: $\log_e \left( \frac{e^x+1}{e^x+5} \right) = - \log_e \left( \frac{e^x+a}{e^x+b} \right)$ Using the logarithm property $-\log X = \log(1/X)$: $\log_e \left( \frac{e^x+1}{e^x+5} \right) = \log_e \left( \left( \frac{e^x+a}{e^x+b} \right)^{-1} \right)$ $\log_e \left( \frac{e^x+1}{e^x+5} \right) = \log_e \left( \frac{e^x+b}{e^x+a} \right)$ Comparing the arguments of the logarithm: $\frac{e^x+1}{e^x+5} = \frac{e^x+b}{e^x+a}$ By comparing the numerators and denominators, we find: $b=1$ $a=5$

Finally, we need to calculate $\frac{|a-b|}{4}$: $\frac{|a-b|}{4} = \frac{|5-1|}{4} = \frac{|4|}{4} = \frac{4}{4} = 1$

The final answer is 1.