The integral \(\int\displaystyle\frac{x^4+2}{x^2+1}\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{3}x^3-x+3\tan^{-1}x+c\), where c is a constant
The numerator has the higher degree, so first divide:
\(x^4+2 = (x^4-1)+3\)
\(= (x^2+1)(x^2-1)+3\)
\( \displaystyle \frac{x^4+2}{x^2+1} = x^2-1+\frac{3}{x^2+1} \)
Integrating term by term, with \(\int\frac{dx}{1+x^2} = \tan^{-1}x\):
\( \displaystyle \int\left(x^2-1+\frac{3}{x^2+1}\right)dx = \frac{1}{3}x^3-x+3\tan^{-1}x+c \)