$\int an^{-1}\sqrt{x}\,dx$ is equal to
(A) $(x+1) an^{-1}\sqrt{x} - \sqrt{x} + C$
(B) $x an^{-1}\sqrt{x} - \sqrt{x} + C$
(C) $\sqrt{x} - x an^{-1}\sqrt{x} + C$
(D) $\sqrt{x} - (x+1) an^{-1}\sqrt{x} + C$

Question

$\int 	an^{-1}\sqrt{x}\,dx$ is equal to
(A) $(x+1)	an^{-1}\sqrt{x} - \sqrt{x} + C$
(B) $x	an^{-1}\sqrt{x} - \sqrt{x} + C$
(C) $\sqrt{x} - x	an^{-1}\sqrt{x} + C$
(D) $\sqrt{x} - (x+1)	an^{-1}\sqrt{x} + C$

Accepted Answer

> [!TLDR]
> The key idea is to integrate $	an^{-1}\sqrt{x}$ by first substituting $t = \sqrt{x}$ to simplify the argument, then applying integration by parts. The correct result is $(x+1)	an^{-1}\sqrt{x} - \sqrt{x} + C$, which matches option **(A)**.

The problem asks for the indefinite integral of $	an^{-1}\sqrt{x}$. The presence of $\sqrt{x}$ inside the inverse tangent makes direct integration by parts messy. A clean approach is to change the variable so that the square root disappears, turning the integrand into something we can handle with standard techniques.

**Why this works:**  
When you see $\sqrt{x}$ inside a function, substituting $t = \sqrt{x}$ (so $x = t^2$) often simplifies the expression. Here, $	an^{-1}\sqrt{x}$ becomes $	an^{-1} t$, and $dx = 2t\,dt$. The integral then becomes $\int 	an^{-1} t \cdot 2t\,dt$, which is a product of a polynomial and an inverse trigonometric function — a classic candidate for integration by parts.

Let’s go step by step.

1. **Substitute $t = \sqrt{x}$.**  
   Then $x = t^2$, so $dx = 2t\,dt$. The integral becomes:
   $$
   I = \int 	an^{-1}\sqrt{x}\,dx = \int 	an^{-1} t \cdot 2t\,dt = 2\int t\,	an^{-1} t\,dt.
   $$

2. **Apply integration by parts.**  
   For $\int t\,	an^{-1} t\,dt$, let:
   $$
   u = 	an^{-1} t \quad 	ext{and} \quad dv = t\,dt.
   $$
   Then:
   $$
   du = \frac{1}{1+t^2}\,dt, \quad v = \frac{t^2}{2}.
   $$
   Integration by parts gives:
   $$
   \int t\,	an^{-1} t\,dt = \frac{t^2}{2}	an^{-1} t - \int \frac{t^2}{2}\cdot\frac{1}{1+t^2}\,dt.
   $$

3. **Simplify the remaining integral.**  
   The integrand $\frac{t^2}{1+t^2}$ can be rewritten:
   $$
   \frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}.
   $$
   So:
   $$
   \int \frac{t^2}{1+t^2}\,dt = \int \left(1 - \frac{1}{1+t^2}ight)dt = t - 	an^{-1} t + C_1.
   $$
   Therefore:
   $$
   \int t\,	an^{-1} t\,dt = \frac{t^2}{2}	an^{-1} t - \frac{1}{2}\left(t - 	an^{-1} tight) + C.
   $$

4. **Multiply by 2 and substitute back.** …

Q. $\int \tan^{-1}\sqrt{x}\,dx$ is equal to
(A) $(x+1)\tan^{-1}\sqrt{x} - \sqrt{x} + C$
(B) $x\tan^{-1}\sqrt{x} - \sqrt{x} + C$
(C) $\sqrt{x} - x\tan^{-1}\sqrt{x} + C$
(D) $\sqrt{x} - (x+1)\tan^{-1}\sqrt{x} + C$

Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Choosing $u$ : the ILATE rule

Worked idea

Method: Substitute away the radical, then integrate by parts

Steps

Common Mistakes

Step 1 — Integration by parts

Step 2 — Complete the integral

Step 3 — Rewrite in the option's form

🎓Unlock everything free for 14 days

Unlock everything free for 14 days