Match List I with List II

List - I	List - II
A. $\int \frac{dx}{x^2 - a^2} =$	I. $\log_e\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$
B. $\int \frac{dx}{a^2 - x^2} =$	II. $\frac{1}{2a}\log_e\left\lvert\frac{x-a}{x+a}\right\rvert + C$
C. $\int \frac{dx}{\sqrt{x^2 - a^2}} =$	III. $\frac{1}{2a}\log_e\left\lvert\frac{a+x}{a-x}\right\rvert + C$
D. $\int \frac{dx}{\sqrt{x^2 + a^2}} =$	IV. $\log_e\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C$

Choose the correct answer from the options given below:

These are standard integral formulas. Let's match them one by one by looking at the structure of each integral.

---

$\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\log_e\left|\dfrac{x-a}{x+a}\right| + C$

The denominator has $x^2 - a^2$ (where $x$ comes first), and the result has $x$ first in the log fraction too.

$A \rightarrow II$

---

$\int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a}\log_e\left|\dfrac{a+x}{a-x}\right| + C$

The denominator has $a^2 - x^2$ (where $a$ comes first), and the result has $a$ first in the log fraction too.

Note: $\dfrac{1}{x^2 - a^2} = -\dfrac{1}{a^2 - x^2}$, so the sign flip swaps the numerator and denominator inside the log. That's why $A$ and $B$ look similar but aren't the same.

$B \rightarrow III$

---

$\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log_e\left|x + \sqrt{x^2 - a^2}\right| + C$

When there's a square root in the denominator, the result is a simple $\log$ (no fraction inside). Whatever is inside the square root ($x^2 - a^2$) stays the same in the answer.

$C \rightarrow IV$

---

$\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \log_e\left|x + \sqrt{x^2 + a^2}\right| + C$

Same structure as $C$, just the sign inside changes from $-$ to $+$.

$D \rightarrow I$

---

Quick way to remember:

— No square root in denominator → answer has $\dfrac{1}{2a}\log\left|\dfrac{\text{something}}{\text{something}}\right|$ (a fraction inside the log)

— Square root in denominator → answer has $\log|x + \sqrt{\text{same expression}}|$ (no fraction inside the log)

---

$A \rightarrow II, \quad B \rightarrow III, \quad C \rightarrow IV, \quad D \rightarrow I$