Question

If $\int \frac{1}{(x+100)\sqrt{x+99}} dx = f(x) + c$ then $f(x) =$

• A. $2\sqrt{x+100}$
• B. $3\sqrt{x+100}$
• C. $2 \tan^{-1}\sqrt{x+99}$
• D. $2 \tan^{-1}\sqrt{x+100}$

Hint:

For integrals of the form $\frac{1}{(ax+b)\sqrt{cx+d}}$, the substitution $u = \sqrt{cx+d}$ usually leads to an arctan or log form.

Answer 1

The Correct Option is C

Step 1: Concept
Use the substitution $u^{2} = x+99$ to simplify the integrand.

Step 2: Meaning
$2u du = dx$. Also, $x+100 = (x+99) + 1 = u^{2} + 1$.

Step 3: Analysis
The integral becomes $\int \frac{2u du}{(u^{2}+1)u} = 2 \int \frac{1}{u^{2}+1} du = 2 \tan^{-1} u$.

Step 4: Conclusion
Substituting $u = \sqrt{x+99}$ back gives $f(x) = 2 \tan^{-1}\sqrt{x+99}$. Final Answer: (C)