Question

\(\int \tan ^{-1}\sqrt{x}dx=\ldots\)

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

Hint:

Substitute x = t^2 and integrate 2t tan^-1 t by parts.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Put \(x=t^{2}\), so \(dx=2t\,dt\). The integral becomes \(\int 2t\tan^{-1}t\,dt\).

Step 2: Integrate by parts
Take \(\tan^{-1}t\) as the first function and \(2t\) as the second:
\(\int 2t\tan^{-1}t\,dt=t^{2}\tan^{-1}t-\int\frac{t^{2}}{1+t^{2}}dt\).

Step 3: Simplify
\(\int\frac{t^{2}}{1+t^{2}}dt=t-\tan^{-1}t\).
So the result is \(t^{2}\tan^{-1}t-t+\tan^{-1}t=(t^{2}+1)\tan^{-1}t-t\).

Step 4: Back-substitute
With \(t=\sqrt{x}\): \((x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+c\).

Final Answer:
The integral is (x + 1) tan inverse of root x minus root x plus c. \[ \boxed{\text{(D) }(x+1)\tan ^{-1}\sqrt{x}-\sqrt{x}+c} \]