Q.Integrate the following function: $\frac{x+2}{\sqrt{x^2 + 2x + 3}}$

Concept: U Substitution — we rewrite the numerator to match the derivative of the radicand, then split the integral.

First, complete the square in the denominator:

$x^2 + 2x + 3 = (x+1)^2 + 2$.

Now rewrite the numerator:

$x+2 = (x+1) + 1$.

The integral becomes

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

For the first term, let $u = (x+1)^2 + 2$, so $du = 2(x+1)\,dx$. This gives $\frac{1}{2} \int u^{-1/2} du = \sqrt{(x+1)^2 + 2}$. …

We start with

The denominator’s radicand is $x^2+2x+3$. Its derivative is $2x+2 = 2(x+1)$. Our numerator is $x+2$, which is close to $x+1$ but not identical. The natural move: split the numerator so that one part is a multiple of the derivative of the radicand (for a direct $u$-substitution), and the leftover is a constant.

1. Rewrite the numerator Write $x+2 = (x+1) + 1$. Then

2. First integral: $u$-substitution Let $u = x^2+2x+3$. Then $du = (2x+2)\,dx = 2(x+1)\,dx$, so $(x+1)\,dx = \frac{du}{2}$. The first integral becomes

Substituting back:

3. Second integral: complete the square The radicand $x^2+2x+3 = (x^2+2x+1) + 2 = (x+1)^2 + 2$. So

This is a standard form: $\displaystyle \int \frac{1}{\sqrt{t^2 + a^2}}\,dt = \sinh^{-1}\!\left(\frac{t}{a}\right) + C$ (or $\log|t + \sqrt{t^2+a^2}| + C$).

Here $t = x+1$ and $a = \sqrt{2}$. Hence

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

Method: $\dfrac{\text{linear}}{\sqrt{\text{quadratic}}}$ with $u^2+a^2$ form

When the completed square gives a sum $u^2+a^2$ (positive definite quadratic), the constant piece integrates to $\log\!\big|u+\sqrt{u^2+a^2}\big|$ (equivalently $\sinh^{-1}$).

Steps

Step 1: Split. With $Q(x)=x^2+bx+c$, write the linear numerator as $\lambda\,Q'(x)+\mu$ where $Q'(x)=2x+b$.

Step 2: Derivative part $\to 2\lambda\sqrt{Q(x)}$. …

Common Mistakes

Mistake 1: Splitting $x+2$ as if $2x+2$ were needed directly.

Why it's wrong: the derivative of $x^2+2x+3$ is $2x+2=2(x+1)$; you should write $x+2=(x+1)+1$ so $(x+1)dx=\tfrac12 du$. Correct approach: match the numerator to $(x+1)$ (half the derivative) plus the leftover constant.

Mistake 2: Choosing the wrong standard form. …