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Evaluate the integral: int(e^(2x)-1)/(e^(2x)+3)dx

Evaluate the integral:\newline∫e2x−1e2x+3dx\int\frac{e^{2x}-1}{e^{2x}+3}dx

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Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. Evaluate the integral:\newline∫e2x−1e2x+3dx\int\frac{e^{2x}-1}{e^{2x}+3}dx
  1. Substitution: Let's do a substitution: let u=e2x+3u = e^{2x} + 3. Then, du=2e2xdxdu = 2e^{2x}dx. We need to solve for dxdx, so dx=du2e2xdx = \frac{du}{2e^{2x}}.
  2. Solve for dx: Substitute uu and dxdx into the integral: ∫e2x−1e2x+3dx=∫u−42u⋅du2e2x\int\frac{e^{2x} - 1}{e^{2x} + 3} dx = \int\frac{u - 4}{2u} \cdot \frac{du}{2e^{2x}}.

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