Question

\( \int \frac{dx}{(x+1)\sqrt{4x+3}} \) is equal to

• A. $\tan^{-1}\sqrt{4x+3} + c$
• B. $3 \tan^{-1}\sqrt{4x+3} + c$
• C. $2 \tan^{-1}\sqrt{4x+3} + c$
• D. $4 \tan^{-1}\sqrt{4x+3} + c$

Hint:

When an integral contains $\sqrt{linear}$, substitute $linear = t^2$ to remove the radical.

Answer 1

The Correct Option is C

Step 1: Substitution
Put $4x+3 = t^{2} \implies 4 dx = 2t dt$. Also, $x = \frac{t^2 - 3}{4}$.
Step 2: Transformation
$I = \int \frac{(1/2) t dt}{(\frac{t^2-3}{4} + 1)t} = \frac{1}{2} \int \frac{dt}{\frac{t^2+1}{4}} = 2 \int \frac{dt}{1+t^{2}}$.
Step 3: Integration
$I = 2 \tan^{-1} t + c = 2 \tan^{-1}\sqrt{4x+3} + c$.
Final Answer: (c)