Q.Integrate the function $\frac{5x+3}{\sqrt{x^2+4x+10}}$

Concept: U Substitution — we rewrite the numerator as a derivative of the denominator’s inside, then split into two standard integrals.

Step 1: Complete the square in the denominator:

$x^2 + 4x + 10 = (x+2)^2 + 6$.

So the integral becomes $\int \frac{5x+3}{\sqrt{(x+2)^2 + 6}} \, dx$.

Step 2: Let $u = x+2$, so $x = u-2$, $dx = du$, and $5x+3 = 5(u-2)+3 = 5u - 7$.

The integral is now $\int \frac{5u - 7}{\sqrt{u^2 + 6}} \, du$.

Step 3: Split:

$5\int \frac{u}{\sqrt{u^2+6}}\,du - 7\int \frac{1}{\sqrt{u^2+6}}\,du$.

First integral: substitute $t = u^2+6$, $dt = 2u\,du$ → $5 \cdot \sqrt{u^2+6}$. …

Why U-Substitution Works Here

When you see a square root of a quadratic in the denominator, your first instinct should be: can I make the numerator match the derivative of the expression inside the square root? That’s because the derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}} \cdot u'$, so if the numerator contains $u'$, the integral collapses into a simple power rule.

Here, the radicand is $x^2+4x+10$. Its derivative is $2x+4$. Our numerator is $5x+3$, which is not exactly $2x+4$, but we can force it to be a linear combination: $5x+3 = A(2x+4) + B$. Solve for $A$ and $B$, and the integral splits into two parts — one a pure u-substitution, the other a standard inverse hyperbolic (or log) form.

1. Set up the split.

   We want $5x+3 = A(2x+4) + B$.

   Expand: $5x+3 = 2A x + 4A + B$.

   Compare coefficients:

   • For $x$: $2A = 5 \implies A = \frac{5}{2}$
   • For constant: $4A + B = 3 \implies 4\cdot\frac{5}{2} + B = 3 \implies 10 + B = 3 \implies B = -7$

   So the integral becomes:

2. First integral: pure u-substitution. Let $u = x^2+4x+10$. Then $du = (2x+4)\,dx$. The first integral becomes:

3. Second integral: complete the square. The denominator’s radicand: $x^2+4x+10 = (x^2+4x+4) + 6 = (x+2)^2 + 6$. So we need:

This is a standard form: $\int \frac{1}{\sqrt{t^2 + a^2}} \, dt = \log\left| t + \sqrt{t^2 + a^2} \right| + C$, where $a^2 = 6$ and $t = x+2$. …

Method: Linear numerator over the square root of a quadratic

Use this whenever the integrand is $\dfrac{px+q}{\sqrt{ax^2+bx+c}}$ — a linear top over the root of a quadratic. The idea is to peel off the part of the numerator that is the derivative of the radicand, leaving a pure "$1/\sqrt{\text{quadratic}}$" piece.

Steps

Step 1: Split the numerator against the radicand's derivative.

Write $px+q = \lambda\,\dfrac{d}{dx}(ax^2+bx+c) + \mu = \lambda(2ax+b) + \mu$, and match the coefficient of $x$ and the constant to solve for $\lambda$ and $\mu$.

Step 2: Integrate the derivative part directly.

since the top is exactly the derivative of what is under the root ($u$-substitution with $u=ax^2+bx+c$). …

Common Mistakes

Mistake 1: Trying a single $u$-substitution for the whole numerator.

Why it's wrong: $u=ax^2+bx+c$ only absorbs the part of the numerator proportional to $2ax+b$; the leftover constant term still has an $x$ hidden in $du$, so a naive substitution leaves a stray $x$. Correct approach: split $px+q=\lambda(2ax+b)+\mu$ first, and handle the $\mu$ piece separately by completing the square.

Mistake 2: Mismatching the coefficients when solving for $\lambda,\mu$.

Why it's wrong: forgetting the leading coefficient $a$ (using $2x+\ldots$ instead of $2ax+b$) gives the wrong $\lambda$ and a derivative piece that doesn't integrate to $2\lambda\sqrt{\text{radicand}}$. Correct approach: always differentiate the actual radicand, then equate coefficients of $x$ and of $1$. …