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GUJCET · Maths · Integrals
\(\int \frac{x^5+1}{x+1} d x=\) __________ \(+c\).
- A \(\sum_{n=1}^4\left((-1)^{n+1} \cdot \frac{x^n}{n}\right)\)
- B \(\sum_{n=1}^4\left((-1)^n \cdot \frac{x^n}{n}\right)\)
- C \(\sum_{n=1}^5\left((-1)^{n+1} \cdot \frac{x^n}{n}\right)\)
- D \(\sum_{n=1}^5\left((-1)^n \cdot \frac{x^n}{n}\right)\)
Answer & Solution
Correct Answer
(C) \(\sum_{n=1}^5\left((-1)^{n+1} \cdot \frac{x^n}{n}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{x^5+1}{x+1} = x^4-x^3+x^2-x+1\) \(\int (x^4-x^3+x^2-x+1) d x = \frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}-\frac{x^2}{2}+x\)
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