Q.Integrate the following function: $\frac{x}{\sqrt{x+4}}$, $x>0$

Concept: U Substitution — we let $u$ equal the expression inside the square root to simplify the integral.

Let $u = x + 4$, so $x = u - 4$ and $dx = du$. The integral becomes:

Integrate term by term:

Substitute back $u = x + 4$:

$\frac{2}{3} (x+4)^{3/2} - 8 \sqrt{x+4} + C$ …

Why substitution works here

When you see a square root of a linear expression like $\sqrt{x+4}$, your first instinct should be to let that whole expression become a new variable. The reason is simple: the derivative of $x+4$ is just $1$, so $dx = du$, and the square root becomes $\sqrt{u}$, which is a clean power $u^{1/2}$. The numerator $x$ is just $u-4$, so the whole integrand becomes a combination of powers of $u$ — and those are the easiest functions to integrate.

Let’s walk through it.

1. Set up the substitution

   Let $u = x + 4$. Then $x = u - 4$, and $dx = du$. The integral becomes:

2. Rewrite as a sum of powers

   Split the fraction:

So the integral is:

3. Integrate term by term

   Using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ for $n \neq -1$:

Putting them together:

4. Substitute back

   Replace $u$ with $x+4$: …

Method: Substitute a linear denominator and split into powers

Use this when $x$ sits over a root of a linear expression, e.g. $\dfrac{x}{\sqrt{x+4}}$: substitute the bracket and rewrite $x$ so every term becomes a power of $u$.

Steps

Step 1: Substitute the inside of the root.

Let $u=x+4$, so $du=dx$ and $x=u-4$.

Step 2: Rewrite as a sum of powers.

$\frac{x}{\sqrt{x+4}}\,dx=\frac{u-4}{u^{1/2}}\,du=(u^{1/2}-4u^{-1/2})\,du.$ …

Common Mistakes

Mistake 1: Not re-expressing $x$ as $u-4$.

Why it's wrong: after $u=x+4$ the numerator $x$ must become $u-4$, otherwise the integral mixes variables. Correct approach: substitute $x=u-4$ so the integrand is $\dfrac{u-4}{u^{1/2}}$.

Mistake 2: Not splitting $\dfrac{u-4}{u^{1/2}}$ into separate powers. …