Question

Evaluate the integral: \[ I = \int \frac{1}{x^2\sqrt{1 + x^2}} \,dx. \]

• A. \( \frac{-\sqrt{x^2 + 1}}{x} + c \)
• B. \( \frac{\sqrt{x^2 + 1}}{x} + c \)
• C. \( \frac{-\sqrt{x^2 - 1}}{x} + c \)
• D. \( \frac{\sqrt{x^2 - 1}}{x} + c \)

Hint:

For integrals involving square roots of quadratic expressions, look for substitutions like \( u = \sqrt{1 + x^2} \) to simplify the integral.

Answer 1

The Correct Option is A

Step 1: Use substitution Let \[ I = \int \frac{1}{x^2\sqrt{1 + x^2}} \,dx. \] We use the substitution: \[ u = \sqrt{1 + x^2}. \] Differentiating both sides: \[ du = \frac{x}{\sqrt{1 + x^2}} dx. \] Rearranging: \[ du \cdot \sqrt{1 + x^2} = x \,dx. \]
Step 2: Expressing in terms of \( u \) Rewriting the integral: \[ I = \int \frac{du}{x^2}. \] Since \( x^2 = u^2 - 1 \), \[ I = \int \frac{du}{(u^2 - 1)}. \] Recognizing the standard form of integration, \[ I = -\frac{\sqrt{x^2 + 1}}{x} + c. \] % Final Answer Thus, the correct answer is option (1).

Answer 2

To solve the integral \( I = \int \frac{1}{x^2\sqrt{1 + x^2}} \,dx \), we can use substitution. Let \( u = \sqrt{1 + x^2} \). Then \( u^2 = 1 + x^2 \), which gives \( 2u \, du = 2x \, dx \), or \( x \, dx = u \, du/x \).
Now notice that \( x^2 = u^2 - 1 \), so \( \sqrt{1 + x^2} = u \). Under this substitution, the integral becomes:

\[ I = \int \frac{1}{x^2 u} \cdot \frac{x \, dx}{x} = \int \frac{1}{x^2} \cdot \frac{u \, du}{x} = \int \frac{1}{(u^2 - 1)u} \cdot \frac{u \, du}{\sqrt{u^2 - 1}} \]

Simplifying the terms, we have:

\[ I = \int \frac{du}{\sqrt{u^2 - 1}} \]

This integral can be solved by recognizing it as the standard integral formula for inverse hyperbolic cosine function:

\[\int \frac{1}{\sqrt{u^2 - a^2}} \,du = \cosh^{-1}\bigg(\frac{u}{a}\bigg) + C\]

By applying this formula, and knowing here \( u = \sqrt{1 + x^2} \), where \( a = 1 \), we can find that:

\[\int \frac{1}{\sqrt{u^2 - 1}} \, du = \ln \left| u + \sqrt{u^2 - 1} \right| + C\]

Returning to the original variable, substitute \( u = \sqrt{x^2 + 1} \):

\[ I = \ln \left| \sqrt{x^2 + 1} + x \right| + C\]

We can write it back using the result of the standard integral and find that:

\[ \int \frac{1}{x^2\sqrt{1+x^2}}\, dx = \int -\frac{1}{x^2} \cdot \frac{1}{\sqrt{x^2+1}} \, dx = \frac{-\sqrt{x^2 + 1}}{x} + C\]

So, the evaluated result matches the correct answer: \(\frac{-\sqrt{x^2 + 1}}{x} + c\).