Question

$\int \frac{x^4+1}{1+x^6}dx=$

• A. $ta{n}^{-1}\left(x\right)-ta{n}^{-1}\left(x^3\right)+c$
• B. $ta{n}^{-1}\left(x\right)-\frac{1}{3}ta{n}^{-1}\left(x^3\right)+c$
• C. $ta{n}^{-1}\left(x\right)+ta{n}^{-1}\left(x^3\right)+c$
• D. $ta{n}^{-1}\left(x\right)+\frac{1}{3}ta{n}^{-1}\left(x^3\right)+c$

Answer 1

The correct option is D $ta{n}^{-1}\left(x\right)+\frac{1}{3}ta{n}^{-1}\left(x^3\right)+c$

$I=\int \frac{x^4+1}{1+x^6}dx=\int \frac{(x^4-x^2+1)+x^2}{(1+x^6)}dx=\int \frac{x^4-x^2+1}{1+x^6}dx+\int \frac{x^2}{1+x^6}dx=\int \frac{1}{1+x^2}dx+\frac{1}{3}\int \frac{3x^2}{1+x^6}dx=ta{n}^{-1}\left(x\right)+\frac{1}{3}ta{n}^{-1}{x}^3+c$