Question

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

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

Hint:

When an integral involves $\sqrt{ax+b}$ in the denominator, try substituting $t=\sqrt{ax+b}$ to simplify it into a standard form.

Answer 1

The Correct Option is C

Step 1: Use substitution: \[ t=\sqrt{x+99} \Rightarrow x=t^2-99,\quad dx=2t\,dt \]
Step 2: Substitute in the integral: \[ \int \frac{dx}{(x+100)\sqrt{x+99}} =\int \frac{2t\,dt}{(t^2-99+100)t} =\int \frac{2\,dt}{t^2+1} \]
Step 3: Integrate using the standard result: \[ \int \frac{dt}{1+t^2}=\tan^{-1}t \]
Step 4: Therefore, \[ f(x)=2\tan^{-1}t =2\tan^{-1}\sqrt{x+99} \]