Q.Evaluate: $\int \dfrac{3x-1}{\sqrt{x^2+9}}\,dx$

Concept: standard integral after splitting the numerator. Here the denominator is $\sqrt{x^2+9}$, so we match the numerator to the derivative of $x^2+9$ (which is $2x$).

Write $3x-1=\dfrac{3}{2}(2x)-1$, so

For the first part put $u=x^2+9$, $du=2x\,dx$: $\dfrac{3}{2}\int u^{-1/2}\,du = \dfrac{3}{2}\cdot 2\sqrt{u}=3\sqrt{x^2+9}$. …

The idea

The denominator carries $\sqrt{x^2+9}=\sqrt{x^2+a^2}$ with $a=3$. Two standard tools cover this kind of integral:

• $\displaystyle\int \frac{f'(x)}{\sqrt{f(x)}}\,dx = 2\sqrt{f(x)}+C$ (a plain substitution), and
• the standard integral $\displaystyle\int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C$.

So we split the numerator $3x-1$ into a part proportional to $2x$ (the derivative of $x^2+9$) and a constant part.

Step 1 — Split the numerator

The derivative of $x^2+9$ is $2x$, so write

Hence

Step 2 — First integral (substitution)

Let $u=x^2+9$, so $du=2x\,dx$:

Multiplying by $\dfrac{3}{2}$ gives $3\sqrt{x^2+9}$.

Step 3 — Second integral (standard form)

With $a=3$, …

Method: Splitting a linear-over-root integral into two standard pieces

Use this for $\int\dfrac{px+q}{\sqrt{\text{quadratic}}}\,dx$. Break the numerator into a part proportional to the derivative of the inside plus a constant part.

Steps

Step 1: Split the numerator.

Write $px+q$ so one part is a multiple of $\tfrac{d}{dx}(\text{quadratic})$ and the rest is constant. For $3x-1$ over $\sqrt{x^2+9}$, note $\tfrac{d}{dx}(x^2+9)=2x$, so $3x=\tfrac32(2x)$.

Step 2: Integrate the "derivative" part.

so the $3x$ piece gives $3\sqrt{x^2+9}$.

Step 3: Integrate the constant part with a standard form. …

Common Mistakes

Mistake 1: Not splitting the numerator.

Why it's wrong: $\dfrac{3x-1}{\sqrt{x^2+9}}$ cannot be integrated in one move; the $3x$ and $-1$ pieces need different standard forms. Correct approach: decompose $3x-1$ into a multiple of $2x$ plus a constant.

Mistake 2: Wrong coefficient on the root term. …