Q.Find $\int \sqrt{x^2 + 2x + 5}\, dx$

Concept: U Substitution – Complete the square inside the square root to match a standard form, then use a trigonometric substitution.

Step 1: Complete the square

$x^2 + 2x + 5 = (x+1)^2 + 4$.

Let $u = x+1$, so $du = dx$. The integral becomes

Step 2: Trigonometric substitution

For $\sqrt{u^2 + a^2}$ with $a=2$, set $u = 2\tan\theta$, $du = 2\sec^2\theta\, d\theta$. Then

The integral is

Step 3: Integrate $\sec^3\theta$

Using the standard reduction:

Thus

Step 4: Back-substitute

$\tan\theta = \frac{u}{2}$, $\sec\theta = \frac{\sqrt{u^2+4}}{2}$. So

and

Therefore

Replace $u = x+1$ and simplify the constant:

Why this approach works

The integral $\int \sqrt{x^2 + 2x + 5}\, dx$ looks like it should be related to $\int \sqrt{u^2 + a^2}\, du$ — a standard form whose answer involves a hyperbolic or trigonometric substitution. But the expression under the square root isn't a simple sum of squares yet; it has a linear term $2x$ that spoils the pattern.

The natural first move is to complete the square. This removes the linear term and reveals the underlying structure: a sum of squares. Once we have $\sqrt{(x+1)^2 + 4}$, the substitution $x+1 = 2\tan\theta$ (or $x+1 = 2\sinh t$) turns the square root into something like $2\sec\theta$, and the $dx$ becomes $2\sec^2\theta\, d\theta$. The integral then becomes a trigonometric integral that we can handle with standard techniques.

Let's walk through it.

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Step-by-step solution

1. Complete the square inside the radical.

We have $x^2 + 2x + 5$. Write it as:

So the integral becomes:

2. Substitute to simplify the variable.

Let $u = x+1$, so $du = dx$. Then:

Now we have the standard form $\int \sqrt{u^2 + a^2}\, du$ with $a = 2$.

3. Choose a trigonometric substitution.

For $\sqrt{u^2 + a^2}$, the standard substitution is $u = a\tan\theta$. Here $a = 2$, so set:

Then:

Since we can restrict $\theta$ to $(-\pi/2, \pi/2)$ where $\sec\theta > 0$, we drop the absolute value: $\sqrt{u^2+4} = 2\sec\theta$.

4. Rewrite the integral in terms of $\theta$.

Substitute everything:

5. Evaluate $\int \sec^3\theta\, d\theta$.

This is a classic integral. Use integration by parts: let $I = \int \sec^3\theta\, d\theta$.

Write $\sec^3\theta = \sec\theta \cdot \sec^2\theta$. Let $dv = \sec^2\theta\, d\theta$, so $v = \tan\theta$, and $u = \sec\theta$, so $du = \sec\theta\tan\theta\, d\theta$.

Then:

Now $\tan^2\theta = \sec^2\theta - 1$, so:

Notice the $\int \sec^3\theta\, d\theta$ appears again — that's our $I$. So:

Bring $I$ to the left:

Thus:

6. Multiply by the constant factor.

Our integral is $4I$, so:

7. Convert back to $u$ (and then $x$).

We have $u = 2\tan\theta$, so $\tan\theta = \frac{u}{2}$. To find $\sec\theta$, use the identity $\sec^2\theta = 1 + \tan^2\theta = 1 + \frac{u^2}{4} = \frac{u^2+4}{4}$. Hence $\sec\theta = \frac{\sqrt{u^2+4}}{2}$ (positive, as before).

Now substitute:

• $\sec\theta\tan\theta = \frac{\sqrt{u^2+4}}{2} \cdot \frac{u}{2} = \frac{u\sqrt{u^2+4}}{4}$.
• So $2\sec\theta\tan\theta = 2 \cdot \frac{u\sqrt{u^2+4}}{4} = \frac{u\sqrt{u^2+4}}{2}$.
• Also $\sec\theta + \tan\theta = \frac{\sqrt{u^2+4}}{2} + \frac{u}{2} = \frac{u + \sqrt{u^2+4}}{2}$.

Thus:

The absolute value inside the log can absorb the constant $2$ in the denominator: $\log\left|\frac{u + \sqrt{u^2+4}}{2}\right| = \log|u + \sqrt{u^2+4}| - \log 2$, and $-\log 2$ is just another constant that merges with $C$. So we write:

8. Replace $u$ with $x+1$.

Finally:

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Method: Complete the Square, Then Standard $\sqrt{u^2+a^2}$ Form

Use this when integrating $\sqrt{\text{quadratic}}$ with a linear term inside: reshape the quadratic into $(x-h)^2+a^2$ so a known formula applies.

Steps

Step 1: Complete the square inside the radical.

Turn $x^2+2x+5$ into $(x+1)^2+4=(x+1)^2+2^2$, so $\sqrt{x^2+2x+5}=\sqrt{(x+1)^2+2^2}$.

Step 2: Apply the standard result for $\sqrt{u^2+a^2}$.

With $u=x+1$, $a=2$, use

Step 3: Back-substitute $u=x+1$.

Replace $u$ and simplify to express everything in $x$, then add $C$:

Common Mistakes

Mistake 1: Not completing the square first.

Why it's wrong: $x^2+2x+5$ isn't a pure $u^2+a^2$ until rewritten as $(x+1)^2+4$. Correct approach: complete the square before choosing a formula.

Mistake 2: Using the $\sqrt{a^2-u^2}$ (arcsine) formula.

Why it's wrong: here the constant term makes $u^2+a^2$ (a sum), which gives a log, not an arcsine. Correct approach: match the sign — a plus needs the $\log$ form.

Mistake 3: Mis-reading $a$ from the completed square.

Why it's wrong: $(x+1)^2+2^2$ means $a=2$ and $\frac{a^2}{2}=2$; using $a=4$ scales the log term wrongly. Correct approach: take $a$ as the square root of the constant.