Q44 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
$\int \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}$ is equal to
- Correct answerA$(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C$
- B$x \tan ^{-1} \sqrt{x}-\sqrt{x}+C$
- C$\sqrt{x}-x \tan ^{-1} \sqrt{x}+C$
- D$\sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+C$
Solution
We have, $I=\int 1 \cdot \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}$
Using by parts,
$\begin{aligned}<br />I & =\tan ^{-1} \sqrt{x} \cdot(x)-\int \frac{1}{1+x} \times \frac{1}{2 \sqrt{x}} \times x d x \\<br />& =x \tan ^{-1} \sqrt{x} \cdot-\int \frac{x}{(1+x) 2 \sqrt{x}} d x \\<br />& =x \tan ^{-1} \sqrt{x}-\int\left(\frac{1+x}{(1+x) 2 \sqrt{x}}-\frac{1}{(1+x) 2 \sqrt{x}}\right) d x \\<br />& =x \tan ^{-1} \sqrt{x}-\int \frac{d x}{2 \sqrt{x}}+\int \frac{d x}{2 \sqrt{x}(1+x)} \\<br />& =x \tan ^{-1} \sqrt{x}-\sqrt{x}+\tan ^{-1} \sqrt{x}+C \\<br />& =(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C<br />\end{aligned}$
Using by parts,
$\begin{aligned}<br />I & =\tan ^{-1} \sqrt{x} \cdot(x)-\int \frac{1}{1+x} \times \frac{1}{2 \sqrt{x}} \times x d x \\<br />& =x \tan ^{-1} \sqrt{x} \cdot-\int \frac{x}{(1+x) 2 \sqrt{x}} d x \\<br />& =x \tan ^{-1} \sqrt{x}-\int\left(\frac{1+x}{(1+x) 2 \sqrt{x}}-\frac{1}{(1+x) 2 \sqrt{x}}\right) d x \\<br />& =x \tan ^{-1} \sqrt{x}-\int \frac{d x}{2 \sqrt{x}}+\int \frac{d x}{2 \sqrt{x}(1+x)} \\<br />& =x \tan ^{-1} \sqrt{x}-\sqrt{x}+\tan ^{-1} \sqrt{x}+C \\<br />& =(x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C<br />\end{aligned}$
Hence, the correct answer is option: 'A'.