Board 2026 Class 12 Maths Q8 Solution: The constant A in ∫ 3ax/(b² + c²x²) dx

The question

If $\displaystyle\int \frac{3ax}{b^2 + c^2x^2}\,dx = A\log\left|b^2 + c^2x^2\right| + K$ ⁠, then the value of A is

(A) $3a$

(B) $\frac{3a}{2b^2}$

(C) $\frac{3a}{b^2c^2}$

(D) $\frac{3a}{2c^2}$

Exam answer

Put $b^2 + c^2x^2 = t$ ⁠. Then $2c^2x\,dx = dt$ ⁠.

Why?

The numerator holds $x\,dx$ ⁠, and the derivative of the denominator is a multiple of $x$ ⁠. That is the sign to put the denominator equal to $t$ ⁠.

$b$ and $c$ are constants, so

\frac{d}{dx}(b^2 + c^2x^2) = 2c^2x

\begin{aligned} &\int \frac{3ax}{b^2 + c^2x^2}\,dx \\ &= \frac{3a}{2c^2}\int \frac{dt}{t} \\ &= \frac{3a}{2c^2}\log|t| + K \\ &= \frac{3a}{2c^2}\log\left|b^2 + c^2x^2\right| + K \end{aligned}

Why?

From $2c^2x\,dx = dt$ ⁠:

\begin{aligned} x\,dx &= \frac{dt}{2c^2} \\ 3ax\,dx &= \frac{3a}{2c^2}\,dt \end{aligned}

Put these into the integral:

\begin{aligned} &\int \frac{3ax\,dx}{b^2 + c^2x^2} \\ &= \int \frac{1}{t} \cdot \frac{3a}{2c^2}\,dt \\ &= \frac{3a}{2c^2}\int \frac{dt}{t} \end{aligned}

Then use $\int \frac{dt}{t} = \log|t|$ ⁠, and put back $t = b^2 + c^2x^2$ ⁠.

$\therefore\ A = \frac{3a}{2c^2}$

Why?

Compare with the given form $A\log\left|b^2 + c^2x^2\right| + K$ ⁠: $A$ is the number in front of the log.

Watch out: $b^2$ is a constant, so its derivative is 0 and it never reaches the answer. That rules out (B) and (C).

Answer: (D) $\dfrac{3a}{2c^2}$

Where this comes from

Chapter 7, Integrals: Integration by substitution.

Question type: Substitution

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Here: The top $3ax$ is a multiple of $2c^2x$ ⁠, the derivative of the bottom $b^2 + c^2x^2$ ⁠. Put the bottom equal to $t$ ⁠, and the integral becomes $\frac{3a}{2c^2}\int\frac{dt}{t}$ ⁠.

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