CUET · MATHS · PYQ PAPER 2025
\(\int \tan ^{-1} \sqrt{x} d x\) equals to : (Here \(C\) is an arbitrary constant)
- 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 \(\sqrt{x}-(x+1) \tan ^{-1} \sqrt{x}+C\)
Answer & Solution
Correct Answer
(A) \((x+1) \tan ^{-1} \sqrt{x}-\sqrt{x}+C\)
Step-by-step Solution
Detailed explanation
Let \(u=\sqrt{x}\). Then \(x=u^2\) and \(dx=2u\,du\). \(\int \tan^{-1} u \cdot 2u \, du\) Using integration by parts \(\int v\,dw = vw - \int w\,dv\) with \(v=\tan^{-1}u\) and \(dw=2u\,du\). \(dv = \frac{1}{1+u^2}\,du\) and \(w=u^2\).…
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