^{-1} {x} \, dx is equal to

Correct Answer is : (x+1)tan⁡−1x−x+C(x+1) ^{-1} {x} - {x} + C(x+1)tan−1x−x+C

Correct Answer is : (x+1)tan⁡−1x−x+C(x+1) ^{-1} {x} - {x} + C(x+1)tan−1x−x+C

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BITSAT 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 104 of 130

Marks: +1, -0

\int \tan^{-1} \sqrt{x} \, dx

Solution: 👈: Video Solution

We have,

I = \int 1 \cdot \tan^{-1} \sqrt{x} \, dx

Using by parts,

I = \tan^{-1} \sqrt{x} \cdot (x) - \int \frac{1}{1+x} \times \frac{1}{2\sqrt{x}} \times x \, dx

= x \tan^{-1} \sqrt{x} \cdot - \int \frac{x}{(1+x)2\sqrt{x}} \, dx

= x \tan^{-1} \sqrt{x} - \int \left( \frac{1+x}{(1+x)2\sqrt{x}} - \frac{1}{(1+x)2\sqrt{x}} \right) dx

= x \tan^{-1} \sqrt{x} - \int \frac{dx}{2\sqrt{x}} + \int \frac{dx}{2\sqrt{x}(1+x)}

= x \tan^{-1} \sqrt{x} - \sqrt{x} + \tan^{-1} \sqrt{x} + C

= (x+1) \tan^{-1} \sqrt{x} - \sqrt{x} + C

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