int frac{x^5+1}{x+1} , dx = . . . . . . + | vedclass

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Question

(C) We know that $x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$.Therefore,the integral becomes $\int \frac{(x+1)(x^4 - x^3 + x^2 - x + 1)}{x+1} \, dx = \i

Vedclass · Accepted Answer

$\sum_{n=1}^5 \left((-1)^{n+1} \cdot \frac{x^n}{n}\right)$

Vedclass · Answer

(C) We know that $x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$.Therefore,the integral becomes $\int \frac{(x+1)(x^4 - x^3 + x^2 - x + 1)}{x+1} \, dx = \int (x^4 - x^3 + x^2 - x + 1) \, dx$.Integrating term by term,we get $\frac{x^5}{5} - \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2} + x + c$.This can be written in summation notation as $\sum_{n=1}^5 (-1)^{n+1} \cdot \frac{x^n}{n} + c$.Comparing this with the given options,the correct option is $C$.

$\int \frac{x^5+1}{x+1} \, dx = $ . . . . . . $+ c$.

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