Question

$\int \frac{x^7+2x^5+x^3+1}{x^2+1}dx=$

• A. $\frac{x^6}{3!}+{\tan }^{-1}x+c$
• B. $\frac{x^5}{5}+\frac{x^3}{3}+{\tan }^{-1}x+c$
• C. $\frac{x^6}{3!}+\frac{x^4}{4}+{\tan }^{-1}x+c$
• D. $\frac{x^6}{3!}+\frac{x^4}{4}+\log (x+1)+c$

Answer 1

The correct option is C $\frac{x^6}{3!}+\frac{x^4}{4}+{\tan }^{-1}x+c$

$\int \frac{x^7+2x^5+x^3+1}{x^2+1}dx=\int \frac{x^5(x^2+1)+x^3(x^2+1)+1}{x^2+1}dx=\int x^{5}dx+\int x^{3}dx+\int \frac{1}{x^2+1}=\frac{x^6}{6}+\frac{x^4}{4}+{\tan }^{-1}x+c$