Show that `I_n+ I_(n-1)=a^n/n`integral `int_0^a t^n/(t+1) dt`

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You need to prove the given reccurence relation `I_n +I_(n-1) = (a^n)/n` .

Supposing that `I_n = int_0^a (t^n)/(t+1) dt` , hence, `I_(n-1) = int_0^a (t^(n-1))/(t+1) dt`

You need to add the integrals such that:

`I_n + I_(n-1) = int_0^a ((t^n)/(t+1) + (t^(n-1))/(t+1)) dt`

Since the fractions have the same denominator, you may write the integrand such that:

`I_n + I_(n-1) = int_0^a (t^n + t^(n-1))/(t+1) dt`

You may factor out `t^n`  such that:

`I_n + I_(n-1) = int_0^a (t^n(1 + t^(-1)))/(t+1) dt`

`I_n + I_(n-1) = int_0^a (t^n(1 + 1/t))/(t+1) dt`

`I_n + I_(n-1) = int_0^a (t^n(t + 1))/(t(t+1)) dt`

You need to reduce by `(t+1)`  such that:

`I_n + I_(n-1) = int_0^a (t^n)/t dt`

`I_n + I_(n-1) = int_0^a t^(n - 1) dt`

`I_n + I_(n-1) = (t^(n - 1 + 1)/(n - 1 + 1))|_0^a`

`I_n + I_(n-1) = (t^n)/n |_0^a => I_n + I_(n-1) = a^n/n - 0^n/n`

`I_n + I_(n-1) = a^n/n`

Hence, evaluating the sum of integrals yields `I_n + I_(n-1) = a^n/n.`

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