Question

$\int \frac{1+x^5}{1+x}dx$ is equal to

• A. $1-x+x^2-x^3+x^4+c$
• B. $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+c$
• C. $(1+x{)}^5+c$
• D. $(1-x{)}^5+c$

Answer 1

The correct option is B $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+c$

Let $I=\int \frac{1+x^5}{1+x}dx$
Now , $1-x+x^2-x^3+x^4=\frac{1+x^5}{1+x}$ [Divide $x^5+1$ by $x+1$]
Hence $I=\int (1-x+x^2-x^3+x^4)dx=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+c$