Q8
1 markMCQSection A

If 3axb2+c2x2dx=Alogb2+c2x2+K\int \frac{3ax}{b^2 + c^2 x^2}\,dx = A \log |b^2 + c^2 x^2| + K, then the value of A is

Integrals
Integration by Substitution

Options

(A)3a3a
(B)3a2b2\frac{3a}{2b^2}
(C)3ab2c2\frac{3a}{b^2 c^2}
(D)3a2c2\frac{3a}{2c^2}
Official Answer

Correct option: (D) 3a2c2\frac{3a}{2c^2}


Substituting u=b2+c2x2 gives du=2c2xdxu = b^2+c^2x^2 \text{ gives } du = 2c^2x\,dx. The integral becomes (3a2c2)duu=(3a2c2)logu+K\left(\frac{3a}{2c^2}\right)\int \frac{du}{u} = \left(\frac{3a}{2c^2}\right)\log|u|+K, so A = 3a2c2\frac{3a}{2c^2}.

integration by substitutionlog integraldu=2c^2x dxindefinite integralconstant of integrationx over quadratic integral

Marking Scheme

  • 11 mark: correct option (D) selected — requires correctly identifying the substitution and its derivative factor.

Hint

Substitute u=b2+c2x2 so that du=2c2xdxu = b^2 + c^2x^2 \text{ so that } du = 2c^2x\,dx, then rewrite the integral purely in terms of u.

Quick Oral Answer

Using u=b2+c2x2, du=2c2xdxu = b^2+c^2x^2,\ du = 2c^2x\,dx, the integral reduces to 3a2c2logu, so A=3a2c2\frac{3a}{2c^2}\log|u|, \text{ so } A = \frac{3a}{2c^2}, option (D).

Analysis & Explanation

This tests the standard substitution technique for integrals of the form xquadratic in x2dx\int \frac{x}{\text{quadratic in } x^2}\,dx.


Setting up the substitution

  • Let u=b2+c2x2u = b^2 + c^2x^2. Then dudx=2c2x, so xdx=du2c2\frac{du}{dx} = 2c^2x, \text{ so } x\,dx = \frac{du}{2c^2}.

Substituting into the integral

  • 3axb2+c2x2dx=3axdxb2+c2x2=3adu/(2c2)u=3a2c2duu\int \frac{3ax}{b^2 + c^2x^2}\,dx = 3a\int \frac{x\,dx}{b^2+c^2x^2} = 3a\int \frac{du/(2c^2)}{u} = \frac{3a}{2c^2}\int \frac{du}{u}.

Integrating

  • 3a2c2duu=3a2c2logu+K=3a2c2logb2+c2x2+K\frac{3a}{2c^2}\int \frac{du}{u} = \frac{3a}{2c^2}\log|u| + K = \frac{3a}{2c^2}\log|b^2+c^2x^2| + K.

Why (D) is correct

  • Comparing with Alogb2+c2x2+KA\log|b^2+c^2x^2|+K, we get A=3a2c2A = \frac{3a}{2c^2}, matching option (D).

Why the other options are wrong

  • (A) 3a ignores the substitution factor entirely, as if du = x dx directly.
  • (B) 3a2b2\frac{3a}{2b^2} mistakenly uses b2b^2 instead of c2c^2 in the denominator, confusing the two constants.
  • (C) 3ab2c2\frac{3a}{b^2c^2} incorrectly combines both constants in the denominator and drops the factor of 2.

Common Mistakes

  1. 1Forgetting to divide by the derivative factor 2c² when substituting u=b2+c2x2u = b^2+c^2x^2, leading to answer (A).
  2. 2Mixing up which constant (b or c) belongs in the denominator, leading to answer (B) or (C).
  3. 3Not simplifying 3a×(12c2)3a \times \left(\frac{1}{2c^2}\right) correctly and leaving an incorrect combined fraction.

Interesting Facts

This substitution pattern — recognizing that the numerator is (a constant times) the derivative of the denominator's variable part — is one of the most frequently tested integration techniques in CBSE board exams.

Integrals of the form f(x)f(x)dx\int \frac{f'(x)}{f(x)}\,dx always reduce to logf(x)+K\log|f(x)|+K, a shortcut that avoids lengthy substitution steps once the pattern is recognized.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

How do you recognize when to use log-form integration?

Whenever the integrand looks like a constant multiple of f(x)/f(x)f'(x)/f(x), i.e., the numerator resembles the derivative of the denominator (up to a constant factor), substituting u=f(x)u = f(x) will reduce the integral to duu=logu+K\int \frac{du}{u} = \log|u|+K.

What is the role of the constant K in this problem?

K is the arbitrary constant of integration, required for any indefinite integral since differentiating a constant gives zero — infinitely many antiderivatives differ only by this constant.