In=∫tn1+t2dt, (n=1,2,3......)⇒I6+I4=

In=∫tn1+t2dt I4+I6=∫t41+t2dt+∫t61+t2dt = ∫t4+t61+t2 dt =∫t4(1+t2)1+t2dt =t55+C

In=∫tn1+t2dt, (n=1,2,3......)⇒I6+I4=

In=∫tn1+t2dt I4+I6=∫t41+t2dt+∫t61+t2dt = ∫t4+t61+t2 dt =∫t4(1+t2)1+t2dt =t55+C

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$I_{n} =$ $\int \frac{t^{n}}{1 + t^{2}} d t, (n = 1, 2, 3......) \Rightarrow I_{6} + I_{4} =$

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$\frac{1}{7} t^{7} + c$

$\frac{1}{4} t^{4} + c$

$\frac{1}{3} t^{3} + c$

$\frac{1}{5} t^{5} + c$

answer is A.

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$I_{n} = \int \frac{t^{n}}{1 + t^{2}} dt I_{4} + I_{6} = \int \frac{t^{4}}{1 + t^{2}} dt + \int \frac{t^{6}}{1 + t^{2}} dt = \int \frac{t^{4} + t^{6}}{1 + t^{2}} dt = \int \frac{t^{4} ({1 + t}^{2})}{(1 + t^{2})} dt = \frac{t^{5}}{5} + C$

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In=∫tn1+t2dt, (n=1,2,3......)⇒I6+I4=

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$I_{n} =$ $\int \frac{t^{n}}{1 + t^{2}} d t, (n = 1, 2, 3......) \Rightarrow I_{6} + I_{4} =$