If I_n = int frac{t^{2n}}{1+t^2} dt,then I_{n | vedclass

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(B) Given $I_n = \int \frac{t^{2n}}{1+t^2} dt$ and $I_{n+1} = \int \frac{t^{2n+2}}{1+t^2} dt$. Consider the difference $I_{n+1} - I_n = \int \frac{t^{

Vedclass · Accepted Answer

$\frac{t^{2n+1}}{2n+1} - I_n$

Vedclass · Answer

(B) Given $I_n = \int \frac{t^{2n}}{1+t^2} dt$ and $I_{n+1} = \int \frac{t^{2n+2}}{1+t^2} dt$. Consider the difference $I_{n+1} - I_n = \int \frac{t^{2n+2} - t^{2n}}{1+t^2} dt$. Factor out $t^{2n}$ from the numerator: $I_{n+1} - I_n = \int \frac{t^{2n}(t^2 - 1)}{1+t^2} dt$. This does not simplify directly. Let us re-evaluate: $I_{n+1} + I_n = \int \frac{t^{2n+2} + t^{2n}}{1+t^2} dt = \int \frac{t^{2n}(t^2 + 1)}{1+t^2} dt = \int t^{2n} dt$. Integrating $t^{2n}$ with respect to $t$,we get $\frac{t^{2n+1}}{2n+1} + C$. Therefore,$I_{n+1} + I_n = \frac{t^{2n+1}}{2n+1} + C$. Thus,$I_{n+1} = \frac{t^{2n+1}}{2n+1} - I_n$.

If $I_n = \int \frac{t^{2n}}{1+t^2} dt$,then $I_{n+1} =$

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