Given F(x) = the integral of dx/(2x + 1) with F(0) = 0, find F((e^2 - 1)/2).

Step-by-step solution

1. 1

   Integrate with the substitution $u=2x+1$. The integrand is a reciprocal of a linear function, so a linear substitution clears it:

   $u=2x+1,\qquad du=2\,dx,\qquad dx=\frac{du}{2}$

   $F(x)=\int\frac{1}{u}\cdot\frac{du}{2}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|2x+1|+C$

   The factor $\tfrac12$ is the derivative of the inside undone; dropping it is the most common error in this family of integrals.

2. 2

   Pin down the constant with the initial condition. Substituting $x=0$:

   $F(0)=\frac{1}{2}\ln|1|+C=0+C$

   and $F(0)=0$ forces

   $C=0\qquad\Longrightarrow\qquad F(x)=\frac{1}{2}\ln|2x+1|$

   The condition is what upgrades a whole family of antiderivatives into a single function.

3. 3

   Simplify the argument before plugging it in. With $x=\dfrac{e^2-1}{2}$,

   $2x+1=2\cdot\frac{e^2-1}{2}+1=(e^2-1)+1=e^2$

   The awkward-looking input was engineered precisely so that $2x+1$ collapses to $e^2$ — always simplify the inside of the logarithm first. Note $e^2>0$, so the absolute value is harmless.

4. 4

   Evaluate. Using $\ln(e^2)=2$:

   $F\!\left(\frac{e^2-1}{2}\right)=\frac{1}{2}\ln\left(e^2\right)=\frac{1}{2}\cdot 2=1$

   $\boxed{1}$

5. 5

   Cross-check as a definite integral. Because $F(0)=0$, the answer is also

   $\int_0^{(e^2-1)/2}\frac{dx}{2x+1}=\left[\frac{1}{2}\ln(2x+1)\right]_0^{(e^2-1)/2}=\frac{1}{2}(2-0)=1$

   Numerically the upper limit is $\approx 3.1945$ and $\tfrac12\ln(2\cdot 3.1945+1)=1.0000$, confirming the value.