Question

If \(\int \frac{dx}{(x^2+9)\sqrt{x^2+16}} = \frac{1}{3\sqrt{7}} \tan^{-1} \left( K \frac{x}{\sqrt{16+x^2}} \right) + c\), then \(K =\)

• A. \(\frac{\sqrt{7}}{3}\)
• B. \(3\sqrt{7}\)
• C. \(\frac{3}{\sqrt{7}}\)
• D. \(\frac{3}{7}\)

Hint:

For integrals involving \(\sqrt{x^2+a^2}\), the substitution \(x = a \tan \theta\) is standard, but substituting the ratio \(u = x/\sqrt{x^2+a^2}\) can often simplify the algebra faster by eliminating square roots immediately.

Answer 1

The Correct Option is A

Step 1: Standard Substitution:
For integrals of type \(\int \frac{dx}{(ax^2+b)\sqrt{cx^2+d}}\), a convenient substitution is \(x = \sqrt{d} \tan \theta\) or substituting \(u = \frac{x}{\sqrt{x^2+16}}\). Let's use \(u = \frac{x}{\sqrt{x^2+16}}\). Squaring: \(u^2 = \frac{x^2}{x^2+16} \implies x^2(1-u^2) = 16u^2 \implies x^2 = \frac{16u^2}{1-u^2}\). Differentiating: \[ du = \frac{\sqrt{x^2+16} - x \frac{x}{\sqrt{x^2+16}}}{x^2+16} dx = \frac{16}{(x^2+16)^{3/2}} dx \] So \(\frac{dx}{\sqrt{x^2+16}} = \frac{x^2+16}{16} du\). From \(x^2 = \frac{16u^2}{1-u^2}\), we have \(x^2+16 = \frac{16u^2 + 16(1-u^2)}{1-u^2} = \frac{16}{1-u^2}\). Thus, \(\frac{dx}{\sqrt{x^2+16}} = \frac{1}{16} \cdot \frac{16}{1-u^2} du = \frac{du}{1-u^2}\).
Step 2: Transform the Integral:
The remaining term is \(\frac{1}{x^2+9}\). \(x^2+9 = \frac{16u^2}{1-u^2} + 9 = \frac{16u^2 + 9 - 9u^2}{1-u^2} = \frac{7u^2+9}{1-u^2}\). So, \(\frac{1}{x^2+9} = \frac{1-u^2}{7u^2+9}\). Integral becomes: \[ I = \int \left( \frac{1-u^2}{7u^2+9} \right) \left( \frac{du}{1-u^2} \right) = \int \frac{du}{7u^2+9} \]
Step 3: Integrate:
\[ I = \int \frac{du}{(\sqrt{7}u)^2 + (3)^2} \] Using formula \(\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})\): \[ I = \frac{1}{\sqrt{7}} \cdot \frac{1}{3} \tan^{-1} \left( \frac{\sqrt{7}u}{3} \right) + c \] \[ I = \frac{1}{3\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{7}}{3} \frac{x}{\sqrt{x^2+16}} \right) + c \]
Step 4: Identify K:
Comparing with the given expression: \[ K = \frac{\sqrt{7}}{3} \]