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\int \frac{1+x^{5}}{1+x}dx=

  • Option 1)

    1-x+x^{2}-x^{3}+x^{4}+c

  • Option 2)

    x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}+c

  • Option 3)

    \left ( 1+x \right )^{5}+c

  • Option 4)

    none

 

Answers (2)

best_answer

As we learnt

Special type of indefinite integration -

Binomial Differential by substitution :

\int x^{m}(a+bx^{n})^{p}dx

- wherein

Where m,n,p are rational numbers 

 

 

\frac{1+x^{5}}{1+x}= \frac{1-\left ( -x \right )^{5}}{1-\left ( -x \right )}

(sum of a G.P. whose first term is 1, common ratio ‘–x’ and number of terms is 5)       

                              = 1-x+x^{2}-x^{3}+x^{4}

\int \frac{1+x^{5}}{1+x}dx=\int \left ( 1-x+x^{2}-x^{3}+x^{4} \right )dx=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}+c

 


Option 1)

1-x+x^{2}-x^{3}+x^{4}+c

Option 2)

x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}+c

Option 3)

\left ( 1+x \right )^{5}+c

Option 4)

none

Posted by

prateek

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Posted by

madanagopal chandrasekhar

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