Question

If \(\mathrm{I}=\int_{0}^{1} \tan^{-1}(\sqrt{x}+1)\,dx\), \(\mathrm{J}=\int_{0}^{1} \frac{\sqrt{x}\,dx}{x+2\sqrt{x}+2}\), then the value of \(2\mathrm{I}+\mathrm{J}\) is equal to :

• A. \(2\tan^{-1}\left(\frac{1}{2}\right)\)
• B. \(2\tan^{-1}(2)-\frac{\pi}{2}\)
• C. \(2\tan^{-1}(2)\)
• D. \(\tan^{-1}(2)\)

Hint:

Put \(x=t^2\) in both integrals and integrate \(I\) by parts; the same rational integral appears in \(2I\) with a minus sign and in \(J\) with a plus sign, so it cancels.

Answer 1

The Correct Option is C

Put \(\sqrt{x}=t\), so \(x=t^2\), \(dx=2t\,dt\) and \(t\) goes from 0 to 1.
\(I=\int_0^1 2t\tan^{-1}(t+1)\,dt\). Integrate by parts with \(u=\tan^{-1}(t+1)\), \(dv=2t\,dt\):
\(I=\left[t^2\tan^{-1}(t+1)\right]_0^1-\int_0^1\frac{t^2}{1+(t+1)^2}\,dt=\tan^{-1}(2)-\int_0^1\frac{t^2}{t^2+2t+2}\,dt\).
Let \(K=\int_0^1\frac{t^2}{t^2+2t+2}\,dt\). Then \(I=\tan^{-1}(2)-K\).
\(J=\int_0^1\frac{t\cdot 2t\,dt}{t^2+2t+2}=2\int_0^1\frac{t^2}{t^2+2t+2}\,dt=2K\).
Therefore \(2I+J=2\tan^{-1}(2)-2K+2K=2\tan^{-1}(2)\).
Answer: (3) \(2\tan^{-1}(2)\)