Let I_{n} = int_{0}^{1} x^{n} tan^{-1} x , dx | vedclass

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(D) We have $I_{n} = \int_{0}^{1} x^{n} 	an^{-1} x \, dx$. Using integration by parts,we get: $I_{n} = \left[ \frac{x^{n+1}}{n+1} 	an^{-1} x ight]

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$a_{1}, a_{2}, a_{3}$ are in $AP$

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(D) We have $I_{n} = \int_{0}^{1} x^{n} 	an^{-1} x \, dx$. Using integration by parts,we get: $I_{n} = \left[ \frac{x^{n+1}}{n+1} 	an^{-1} x ight]_{0}^{1} - \int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{1}{1+x^{2}} \, dx$ $I_{n} = \frac{\pi}{4(n+1)} - \frac{1}{n+1} \int_{0}^{1} \frac{x^{n+1}}{1+x^{2}} \, dx$. Now,consider $I_{n} + I_{n+2} = \int_{0}^{1} x^{n} (1+x^{2}) 	an^{-1} x \, dx$. This does not simplify directly,so we use the recurrence relation derived from $I_{n+2} + I_{n} = \int_{0}^{1} x^{n}(1+x^{2}) 	an^{-1} x \, dx$. Alternatively,using the reduction formula: $(n+3) I_{n+2} + (n+1) I_{n} = \frac{\pi}{2} - \frac{1}{n+2}$. Comparing this with $a_{n} I_{n+2} + b_{n} I_{n} = c_{n}$,we get $a_{n} = n+3$ and $b_{n} = n+1$. Since $a_{n} = n+3$,$a_{1}=4, a_{2}=5, a_{3}=6$,which are in $AP$. Since $b_{n} = n+1$,$b_{1}=2, b_{2}=3, b_{3}=4$,which are in $AP$. Thus,both $a_{n}$ and $b_{n}$ are in $AP$.

Let $I_{n} = \int_{0}^{1} x^{n} \tan^{-1} x \, dx$. If $a_{n} I_{n+2} + b_{n} I_{n} = c_{n}$ for all $n \geq 1$,then

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