Skip to content
Exercise 7.2 · Q19

Q.Integrate the following function: e2x−1e2x+1\frac{e^{2x}-1}{e^{2x}+1}

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
Appeared in past exams:GUJCET 2024· Set 13· 1mexact
12% · 45/373 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Dividing top and bottom by exe^{x} makes the numerator the exact derivative of the denominator, so the integral is a ∫duu\int \tfrac{du}{u} log form: log⁡(ex+e−x)+C\log\left(e^{x}+e^{-x}\right)+C.

The idea

Whenever an integrand looks like g′(x)g(x)\dfrac{g'(x)}{g(x)}, the answer is log⁡∣g(x)∣\log|g(x)|. The given fraction isn't in that shape yet, but a single algebraic step puts it there.

Step 1 — Rewrite the integrand

Divide numerator and denominator by exe^{x}:

e2x−1e2x+1=e2x ⁣⋅ ⁣e−x−e−xe2x ⁣⋅ ⁣e−x+e−x=ex−e−xex+e−x.\frac{e^{2x}-1}{e^{2x}+1} = \frac{e^{2x}\!\cdot\! e^{-x}-e^{-x}}{e^{2x}\!\cdot\! e^{-x}+e^{-x}} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}.

Step 2 — Choose the substitution

Let u=ex+e−xu=e^{x}+e^{-x} (the denominator). Then

dudx=ex−e−x⇒du=(ex−e−x) dx,\frac{du}{dx}=e^{x}-e^{-x} \quad\Rightarrow\quad du=(e^{x}-e^{-x})\,dx,

which is precisely the numerator times dxdx.

Step 3 — Integrate

∫ex−e−xex+e−x dx=∫duu=log⁡∣u∣+C.\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx = \int \frac{du}{u} = \log|u| + C.

Step 4 — Back-substitute …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.