Q.Integrate the function $\sqrt{ax+b}$

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

Let $u = ax + b$. Then $du = a\,dx$, so $dx = \frac{du}{a}$.

The integral becomes:

Integrate: …

Why U-Substitution Works Here

When you see a composite function like $\sqrt{ax+b}$, your first instinct should be to look for an "inner function" that you can replace with a single variable. The square root is an "outer" operation applied to the linear expression $ax+b$. If we let $u = ax+b$, then the square root becomes $\sqrt{u}$, which is just $u^{1/2}$ — a simple power function we know how to integrate.

The real magic is that the derivative of $ax+b$ is a constant $a$, so $du = a\,dx$ gives us a clean way to replace $dx$ as well. No messy chain rule to untangle — just a straight substitution.

Step-by-Step Solution

1. Choose the substitution.

Let $u = ax + b$. This is the expression inside the square root. The entire integral is $\int \sqrt{ax+b}\,dx$, so replacing $ax+b$ with $u$ gives $\int \sqrt{u}\,dx$ — but we still have $dx$, not $du$.

2. Find $du$ in terms of $dx$.

Differentiate $u$ with respect to $x$:

So $du = a\,dx$, which means $dx = \frac{du}{a}$.

3. Rewrite the integral entirely in $u$.

Substitute $dx$ and the square root:

$\int \sqrt{ax+b}\,dx = \int \sqrt{u} \cdot \frac{du}{a} = \frac{1}{a} \int u^{1/2}\,du$ …

Method: Substitution for a Linear Inner Function

Use this whenever a single linear expression $ax+b$ sits inside another operation (a root, a power, an exponential): replace the whole linear chunk with one new variable.

Steps

Step 1: Set $u$ equal to the linear expression.

Let $u=ax+b$. Then $du=a\,dx$, so $dx=\dfrac{du}{a}$. Because the derivative of a linear expression is the constant $a$, the swap is clean — no leftover $x$ terms remain.

Step 2: Rewrite the whole integral in $u$. …

Common Mistakes

Mistake 1: Forgetting the $\dfrac1a$ factor from $dx=\frac{du}{a}$.

Why it's wrong: substitution changes $dx$, and skipping the $\frac1a$ scales the whole answer wrongly. Correct approach: always compute $du$ and solve for $dx$ before substituting.

Mistake 2: Leaving the answer in terms of $u$. …