Draw the Lewis structure and 3D shape/structure (with perspective) for IBr4*. State the electron pair geometry and the molecular geometry for your structure in #9. What is/are the bond angle(s) in your structure in #9?
Make a substitution to express the integrand as a rational function and then evaluate the integral. (Use C for the constant of integration.)
The Correct Answer and Explanation is:
We are given the integral: ∫2e2xe2x+16ex+63 dx\int \frac{2e^{2x}}{e^{2x} + 16e^x + 63} \, dx
Step 1: Substitution
Let us simplify the expression by substituting: u=ex⇒du=exdx⇒dx=duuu = e^x \Rightarrow du = e^x dx \Rightarrow dx = \frac{du}{u}
Then: e2x=(ex)2=u2e^{2x} = (e^x)^2 = u^2
Now rewrite the integral: ∫2e2xe2x+16ex+63 dx=∫2u2u2+16u+63⋅1u du=∫2uu2+16u+63 du\int \frac{2e^{2x}}{e^{2x} + 16e^x + 63} \, dx = \int \frac{2u^2}{u^2 + 16u + 63} \cdot \frac{1}{u} \, du = \int \frac{2u}{u^2 + 16u + 63} \, du
Step 2: Use Substitution for Rational Function
We will now use a direct substitution to integrate the rational function:
Let the denominator be: u2+16u+63u^2 + 16u + 63
We complete the square: u2+16u+63=(u+8)2−1u^2 + 16u + 63 = (u + 8)^2 – 1
But since this is a simple rational function, we can integrate using u-substitution directly.
Let: v=u2+16u+63⇒dvdu=2u⇒du=dv2uv = u^2 + 16u + 63 \Rightarrow \frac{dv}{du} = 2u \Rightarrow du = \frac{dv}{2u}
Then: ∫2uv du=∫2uv⋅dv2u=∫dvv=ln∣v∣+C=ln∣u2+16u+63∣+C\int \frac{2u}{v} \, du = \int \frac{2u}{v} \cdot \frac{dv}{2u} = \int \frac{dv}{v} = \ln|v| + C = \ln|u^2 + 16u + 63| + C
Now recall that u=exu = e^x, so: u2+16u+63=e2x+16ex+63u^2 + 16u + 63 = e^{2x} + 16e^x + 63
✅ Final Answer:
∫2e2xe2x+16ex+63 dx=ln∣e2x+16ex+63∣+C\boxed{\int \frac{2e^{2x}}{e^{2x} + 16e^x + 63} \, dx = \ln|e^{2x} + 16e^x + 63| + C}
📘 Explanation
This integral involves an exponential expression in both the numerator and denominator. To simplify, we use substitution. Setting u=exu = e^x converts the exponential terms into polynomials, allowing us to express the integrand as a rational function. This substitution gives e2x=u2e^{2x} = u^2 and dx=duudx = \frac{du}{u}. Substituting into the integral: 2e2xe2x+16ex+63dx=2u2u2+16u+63⋅1u du=2uu2+16u+63 du\frac{2e^{2x}}{e^{2x} + 16e^x + 63} dx = \frac{2u^2}{u^2 + 16u + 63} \cdot \frac{1}{u} \, du = \frac{2u}{u^2 + 16u + 63} \, du
This is now a rational function where the numerator is the derivative (up to a constant) of the denominator. Recognizing this structure allows us to apply a substitution v=u2+16u+63v = u^2 + 16u + 63. Then, dv=(2u+16)dudv = (2u + 16)du, but since we only have 2u du2u\,du, we simply do: dv=2u du⇒dv2u=dudv = 2u \, du \Rightarrow \frac{dv}{2u} = du
Plugging in, the integral becomes: ∫2uv⋅du=∫dvv=ln∣v∣+C\int \frac{2u}{v} \cdot du = \int \frac{dv}{v} = \ln|v| + C
Back-substituting v=u2+16u+63v = u^2 + 16u + 63, and u=exu = e^x, we get the final answer: ln∣e2x+16ex+63∣+C\ln|e^{2x} + 16e^x + 63| + C
This method shows the power of substitution in simplifying complex-looking exponential integrals.
