Chapter 5: Q. 49 (page 465)

Solve the integral
$\int \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$

Short Answer

Expert verified

$\frac{\sqrt{x - 3}}{9 x} \sqrt{x + 3} + C$

Step by step solution

Step 1. Given

$\int \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$

Step 2. Using identity

$To solve the integral, let x = 3 \cosh (u) . Now, derivation of the above equation is solved below . x = 3 \cos h (u) d x = 3 \sin h ((u) d u) Now, use the identity c o s h^{2} (u) - 1 = s i n h^{2} (u) S o, \frac{1}{x^{2} \sqrt{x^{2} - 9}} = \frac{1}{(9 \sqrt[]{(9 \cos h^{2} (u) - 9)} c o s h^{2} (u))} = \frac{1}{27 \sqrt{\cos h^{2} (u)} - 1 c o s h^{2} (u)} = \frac{1}{27 \sqrt{s i n h^{2} (u}) c o s h^{2} (u)} = \frac{1}{27 s i n h (u) c o s h^{2} (u)}$

Step 3. Calculation

$\frac{1}{x^{2} \sqrt{x^{2} - 9}} = \int \frac{1}{9 c o s h^{2} (u)} d u = \frac{1}{t a n h (u)} + C = \frac{1}{9 t a n h (c o s h^{-} 1 (\frac{x}{2}))} + C = \frac{\sqrt{\frac{x}{3} + 1}}{3 x} \sqrt{\frac{x}{3} - 1} + C = \frac{\sqrt{x - 3}}{9 x} \sqrt{x + 3} + C Therefore, the value is boxed \frac{\sqrt{x - 3}}{9 x} \sqrt{x + 3} + C$

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Solve the integral
$\int \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$

Short Answer

Step by step solution

Step 1. Given

Step 2. Using identity

Step 3. Calculation

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Most popular questions from this chapter

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Solve the integral
$\int \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$