Question

Evaluate \(\rm \int\sqrt{\frac{1+x}{1-x}}\ dx\), x ≠ 1.
1. \(\rm \cos^{-1}x+\sqrt{1-x^2}+C\).
2. \(\rm \tan^{-1}x+\sqrt{1-x^2}+C\).
3. \(\rm \sin^{-1}x+\sqrt{1-x^2}+C\).
4. None of these.

Answer 1

Correct Answer - Option 4 : None of these.

Concept:

• Integration by substitution: If we substitute x = f(t), then dx = f'(t) dt and ∫ f(x) dx = ∫ f[f(t)] f'(t) dt.
• \(\rm \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C\)..

 

Calculation:

Let I = \(\rm \int\sqrt{\frac{1+x}{1-x}}\ dx\)

I = \(\rm \int\sqrt{\frac{1+x}{1-x} \times \frac{1+x}{1+x} }\ dx\)

I = \(\rm \int\frac{1+x}{\sqrt{1-x^2}}\ \ dx\)

⇒ I = \(\rm \int\frac{1}{\sqrt{1-x^2}}\ dx+\int\frac{x}{\sqrt{1-x^2}}\ dx\)

⇒ I = sin^-1 x + I₁

Where I₁ = \(\rm \int\frac{x}{\sqrt{1-x^2}}\ dx\).

Substituting 1 - x² = t²,

⇒ -(2x)dx = 2tdt

⇒ xdx = -tdt

⇒ I₁ = - ∫ dt

= -t + C

= -\(\rm \sqrt{1-x^2}\) + C

Therefore, I = sin-1 x - \(\rm \sqrt{1-x^2}\) + C.

Hence, the correct answer is None of these.