Q.$\int \sqrt{1+x^2} dx$ is equal to (A) $\frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log |x+\sqrt{1+x^2}| + C$ (B) $\frac{2}{3}(1+x^2)^{\frac{3}{2}} + C$ (C) $\frac{2}{3}x(1+x^2)^{\frac{3}{2}} + C$ (D) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log |x+\sqrt{1+x^2}| + C$

Concept: Integration by parts (or the standard formula for $\int \sqrt{a^2 + x^2}\,dx$).

Step 1: Write $\sqrt{1+x^2} = 1 \cdot \sqrt{1+x^2}$ and integrate by parts:

Let $u = \sqrt{1+x^2}$, $dv = dx$. Then $du = \frac{x}{\sqrt{1+x^2}}\,dx$, $v = x$.

Step 2:

$\int \sqrt{1+x^2}\,dx = x\sqrt{1+x^2} - \int \frac{x^2}{\sqrt{1+x^2}}\,dx$.

Step 3: Add and subtract 1 in the numerator:

$\int \frac{x^2}{\sqrt{1+x^2}}\,dx = \int \frac{(x^2+1)-1}{\sqrt{1+x^2}}\,dx = \int \sqrt{1+x^2}\,dx - \int \frac{dx}{\sqrt{1+x^2}}$.

Step 4: Substitute back and solve for the original integral:

$\int \sqrt{1+x^2}\,dx = x\sqrt{1+x^2} - \left( \int \sqrt{1+x^2}\,dx - \int \frac{dx}{\sqrt{1+x^2}} \right)$ …

The key insight: when you see $\sqrt{a^2 + x^2}$, your first thought might be a trigonometric substitution ($x = \tan\theta$). That works, but there's an elegant algebraic trick that avoids trig entirely — integration by parts, treating the integrand as $1 \cdot \sqrt{1+x^2}$.

Why does this work? Because differentiating $\sqrt{1+x^2}$ gives $\frac{x}{\sqrt{1+x^2}}$, and integrating $1$ gives $x$. The product rule in reverse then produces a new integral that simplifies beautifully when you add and subtract the right term.

Let's walk through it.

1. Set up integration by parts. Let $u = \sqrt{1+x^2}$ and $dv = dx$. Then $du = \frac{x}{\sqrt{1+x^2}}\,dx$ and $v = x$. Integration by parts gives:

2. The trick: add and subtract $1$ in the numerator. The new integral looks almost like the original, but with $x^2$ instead of $1+x^2$. Notice:

So the integral becomes:

3. Separate and bring like terms together.

Now add $\int \sqrt{1+x^2}\,dx$ to both sides:

4. The remaining integral is a standard form. $\int \frac{1}{\sqrt{1+x^2}}\,dx = \sinh^{-1} x + C = \log |x + \sqrt{1+x^2}| + C$ …

Method: Recognise the standard $\int\sqrt{1+x^{2}}\,dx$ result

For an MCQ asking for $\int\sqrt{x^{2}+a^{2}}\,dx$, match against the memorised standard form rather than re-deriving under time pressure.

Steps

Step 1: Identify the form. $\sqrt{1+x^{2}}=\sqrt{x^{2}+a^{2}}$ with $a^{2}=1$.

Step 2: Recall the template.

Step 3: Substitute $a^{2}=1$. …

Common Mistakes

Mistake 1: Treating $\sqrt{1+x^{2}}$ like $(1+x^{2})^{1/2}$ and using the power rule.

Why it's wrong: the power rule $\int u^{n}du$ needs $du$ present; here it gives the wrong $\tfrac23(1+x^{2})^{3/2}$ distractor (B). Correct approach: use the dedicated $\sqrt{x^{2}+a^{2}}$ formula.

Mistake 2: Expecting no logarithm in the answer. …