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Question: Which of the following option(s) is/are correct (where C is constant of integration)...

Which of the following option(s) is/are correct (where C is constant of integration)

A

∫e2x−1e2x+1dx=ln⁡(1+e2x)−x+C\int \frac{e^{2x}-1}{e^{2x}+1}dx = \ln(1+e^{2x})-x+C

B

∫e−x1+exdx=ln⁡(1+ex)−x−ex+C\int \frac{e^{-x}}{1+e^{x}}dx = \ln(1+e^{x})-x-e^{x}+C

C

∫1(ex+e−x)2dx=−12(1+e2x)+C\int \frac{1}{(e^{x}+e^{-x})^2}dx = -\frac{1}{2(1+e^{2x})}+C

D

∫11−e2xdx=ln⁡(1+1−e2xex)+C\int \frac{1}{\sqrt{1-e^{2x}}}dx = \ln(\frac{1+\sqrt{1-e^{2x}}}{e^{x}})+C

Answer

A, C

Explanation

Solution

To determine which of the given options are correct, we will evaluate each integral or differentiate the proposed solution and check if it matches the integrand.

Option A: ∫e2x−1e2x+1dx=ln⁡(1+e2x)−x+C\int \frac{e^{2x}-1}{e^{2x}+1}dx = \ln(1+e^{2x})-x+C

Let's differentiate the right-hand side (RHS): ddx(ln⁡(1+e2x)−x+C)\frac{d}{dx} \left( \ln(1+e^{2x})-x+C \right) =11+e2x⋅ddx(1+e2x)−1+0= \frac{1}{1+e^{2x}} \cdot \frac{d}{dx}(1+e^{2x}) - 1 + 0 =11+e2x⋅(2e2x)−1= \frac{1}{1+e^{2x}} \cdot (2e^{2x}) - 1 =2e2x1+e2x−1+e2x1+e2x= \frac{2e^{2x}}{1+e^{2x}} - \frac{1+e^{2x}}{1+e^{2x}} =2e2x−(1+e2x)1+e2x= \frac{2e^{2x} - (1+e^{2x})}{1+e^{2x}} =2e2x−1−e2x1+e2x= \frac{2e^{2x} - 1 - e^{2x}}{1+e^{2x}} =e2x−1e2x+1= \frac{e^{2x}-1}{e^{2x}+1} This matches the integrand. So, Option A is correct.

Alternatively, to integrate ∫e2x−1e2x+1dx\int \frac{e^{2x}-1}{e^{2x}+1}dx: Divide the numerator and denominator by exe^x: ∫ex−e−xex+e−xdx\int \frac{e^x - e^{-x}}{e^x + e^{-x}}dx Let u=ex+e−xu = e^x + e^{-x}. Then du=(ex−e−x)dxdu = (e^x - e^{-x})dx. The integral becomes ∫duu=ln⁡∣u∣+C=ln⁡(ex+e−x)+C\int \frac{du}{u} = \ln|u| + C = \ln(e^x + e^{-x}) + C. We can rewrite ln⁡(ex+e−x)\ln(e^x + e^{-x}) as ln⁡(e2x+1ex)=ln⁡(e2x+1)−ln⁡(ex)=ln⁡(1+e2x)−x\ln(\frac{e^{2x}+1}{e^x}) = \ln(e^{2x}+1) - \ln(e^x) = \ln(1+e^{2x}) - x. So, the integral is ln⁡(1+e2x)−x+C\ln(1+e^{2x}) - x + C. This confirms Option A is correct.

Option B: ∫e−x1+exdx=ln⁡(1+ex)−x−ex+C\int \frac{e^{-x}}{1+e^{x}}dx = \ln(1+e^{x})-x-e^{x}+C

Let's differentiate the RHS: ddx(ln⁡(1+ex)−x−ex+C)\frac{d}{dx} \left( \ln(1+e^{x})-x-e^{x}+C \right) =11+ex⋅ddx(1+ex)−1−ex+0= \frac{1}{1+e^{x}} \cdot \frac{d}{dx}(1+e^{x}) - 1 - e^{x} + 0 =ex1+ex−1−ex= \frac{e^{x}}{1+e^{x}} - 1 - e^{x} =ex−(1+ex)−ex(1+ex)1+ex= \frac{e^{x} - (1+e^{x}) - e^{x}(1+e^{x})}{1+e^{x}} =ex−1−ex−ex−e2x1+ex= \frac{e^{x} - 1 - e^{x} - e^{x} - e^{2x}}{1+e^{x}} =−1−ex−e2x1+ex= \frac{-1 - e^{x} - e^{2x}}{1+e^{x}} The integrand is e−x1+ex=1ex(1+ex)\frac{e^{-x}}{1+e^{x}} = \frac{1}{e^x(1+e^x)}. Since −1−ex−e2x1+ex≠e−x1+ex\frac{-1 - e^{x} - e^{2x}}{1+e^{x}} \neq \frac{e^{-x}}{1+e^{x}}, Option B is incorrect.

Option C: ∫1(ex+e−x)2dx=−12(1+e2x)+C\int \frac{1}{(e^{x}+e^{-x})^2}dx = -\frac{1}{2(1+e^{2x})}+C

Let's differentiate the RHS: ddx(−12(1+e2x)+C)\frac{d}{dx} \left( -\frac{1}{2(1+e^{2x})}+C \right) =−12ddx(1+e2x)−1= -\frac{1}{2} \frac{d}{dx} (1+e^{2x})^{-1} =−12(−1)(1+e2x)−2⋅ddx(1+e2x)= -\frac{1}{2} (-1) (1+e^{2x})^{-2} \cdot \frac{d}{dx}(1+e^{2x}) =121(1+e2x)2⋅(2e2x)= \frac{1}{2} \frac{1}{(1+e^{2x})^2} \cdot (2e^{2x}) =e2x(1+e2x)2= \frac{e^{2x}}{(1+e^{2x})^2} Now let's simplify the integrand: 1(ex+e−x)2=1(e2x+1ex)2=1(e2x+1)2(ex)2=e2x(e2x+1)2\frac{1}{(e^{x}+e^{-x})^2} = \frac{1}{\left(\frac{e^{2x}+1}{e^x}\right)^2} = \frac{1}{\frac{(e^{2x}+1)^2}{(e^x)^2}} = \frac{e^{2x}}{(e^{2x}+1)^2} This matches the derivative of the RHS. So, Option C is correct.

Option D: ∫11−e2xdx=ln⁡(1+1−e2xex)+C\int \frac{1}{\sqrt{1-e^{2x}}}dx = \ln(\frac{1+\sqrt{1-e^{2x}}}{e^{x}})+C

Let's differentiate the RHS: Let y=ln⁡(1+1−e2xex)=ln⁡(1+1−e2x)−ln⁡(ex)=ln⁡(1+1−e2x)−xy = \ln\left(\frac{1+\sqrt{1-e^{2x}}}{e^{x}}\right) = \ln(1+\sqrt{1-e^{2x}}) - \ln(e^x) = \ln(1+\sqrt{1-e^{2x}}) - x. dydx=11+1−e2x⋅ddx(1−e2x)−1\frac{dy}{dx} = \frac{1}{1+\sqrt{1-e^{2x}}} \cdot \frac{d}{dx}(\sqrt{1-e^{2x}}) - 1 =11+1−e2x⋅121−e2x⋅ddx(1−e2x)−1= \frac{1}{1+\sqrt{1-e^{2x}}} \cdot \frac{1}{2\sqrt{1-e^{2x}}} \cdot \frac{d}{dx}(1-e^{2x}) - 1 =11+1−e2x⋅121−e2x⋅(−2e2x)−1= \frac{1}{1+\sqrt{1-e^{2x}}} \cdot \frac{1}{2\sqrt{1-e^{2x}}} \cdot (-2e^{2x}) - 1 =−e2x1−e2x(1+1−e2x)−1= \frac{-e^{2x}}{\sqrt{1-e^{2x}}(1+\sqrt{1-e^{2x}})} - 1 =−e2x−1−e2x(1+1−e2x)1−e2x(1+1−e2x)= \frac{-e^{2x} - \sqrt{1-e^{2x}}(1+\sqrt{1-e^{2x}})}{\sqrt{1-e^{2x}}(1+\sqrt{1-e^{2x}})} =−e2x−1−e2x−(1−e2x)1−e2x(1+1−e2x)= \frac{-e^{2x} - \sqrt{1-e^{2x}} - (1-e^{2x})}{\sqrt{1-e^{2x}}(1+\sqrt{1-e^{2x}})} =−e2x−1−e2x−1+e2x1−e2x(1+1−e2x)= \frac{-e^{2x} - \sqrt{1-e^{2x}} - 1 + e^{2x}}{\sqrt{1-e^{2x}}(1+\sqrt{1-e^{2x}})} =−(1+1−e2x)1−e2x(1+1−e2x)= \frac{-(1+\sqrt{1-e^{2x}})}{\sqrt{1-e^{2x}}(1+\sqrt{1-e^{2x}})} =−11−e2x= -\frac{1}{\sqrt{1-e^{2x}}} The derivative is −11−e2x-\frac{1}{\sqrt{1-e^{2x}}}, which is the negative of the integrand 11−e2x\frac{1}{\sqrt{1-e^{2x}}}. So, Option D is incorrect.

The correct options are A and C.

Explanation of the Solution: For each option, the proposed integral solution was verified by differentiating it. If the derivative matched the integrand, the option was marked as correct.

  • For Option A, differentiating ln⁡(1+e2x)−x+C\ln(1+e^{2x})-x+C yielded e2x−1e2x+1\frac{e^{2x}-1}{e^{2x}+1}, matching the integrand.
  • For Option B, differentiating ln⁡(1+ex)−x−ex+C\ln(1+e^{x})-x-e^{x}+C did not yield e−x1+ex\frac{e^{-x}}{1+e^{x}}.
  • For Option C, differentiating −12(1+e2x)+C-\frac{1}{2(1+e^{2x})}+C yielded e2x(1+e2x)2\frac{e^{2x}}{(1+e^{2x})^2}, which is equivalent to the integrand 1(ex+e−x)2\frac{1}{(e^{x}+e^{-x})^2}.
  • For Option D, differentiating ln⁡(1+1−e2xex)+C\ln(\frac{1+\sqrt{1-e^{2x}}}{e^{x}})+C yielded −11−e2x-\frac{1}{\sqrt{1-e^{2x}}}, which is the negative of the integrand.