Question

\[ \int \frac{dx}{x\sqrt{x^2+4}}= \]

• A. \[ \frac12\log\left| \frac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}-2} \right|+C \]
• B. \[ \frac14\log\left| \frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2} \right|+C \]
• C. \[ \frac12\log\left| \frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2} \right|+C \]
• D. \[ \frac14\log\left| \frac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}-2} \right|+C \]

Hint:

Whenever radicals like $\sqrt{x^2+a^2}$ appear, try substituting the entire radical as a new variable. It often converts the integral into a standard rational form.

Answer 1

The Correct Option is B

Concept: Integrals involving expressions of the form: \[ \sqrt{x^2+a^2} \] are usually solved using substitutions such as: \[ x=a\tan\theta \] or by converting the integral into the standard form: \[ \int\frac{du}{u^2-a^2} \] Here, we use a substitution involving the radical itself.

Step 1: Introduce a suitable substitution.
Let \[ u=\sqrt{x^2+4} \] Squaring both sides: \[ u^2=x^2+4 \] Differentiating: \[ 2u\,du=2x\,dx \] Therefore, \[ u\,du=x\,dx \] or \[ dx=\frac{u}{x}\,du \]

Step 2: Transform the integral.
The integral becomes: \[ I=\int \frac{1}{x\,u}\cdot\frac{u}{x}\,du \] \[ I=\int \frac{du}{x^2} \] From \[ u^2=x^2+4 \] we get: \[ x^2=u^2-4 \] Hence, \[ I=\int\frac{du}{u^2-4} \]

Step 3: Use the standard integral formula.
Recall: \[ \int\frac{du}{u^2-a^2} = \frac{1}{2a} \log\left| \frac{u-a}{u+a} \right|+C \] Here, \[ a=2 \] Therefore, \[ I= \frac14 \log\left| \frac{u-2}{u+2} \right|+C \] Substituting back: \[ u=\sqrt{x^2+4} \] Thus, \[ I= \frac14 \log\left| \frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2} \right|+C \] Hence, \[ \boxed{ \frac14 \log\left| \frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2} \right|+C } \]