∫ x^4+1 / x^2+1 dx =

Question

\(\int \frac{x^{4}+1}{x^{2}+1} \mathrm{~d} x=\)

(A) \(\frac{x^{3}}{3}+c\)

(B) \( \frac {x^3}3 - x + \tan^{-1} x + c\)

(C) \(2 \tan ^{-1} x+c\)

(D) \(\frac{x^{3}}{3}+x+2 \tan ^{-1} x+c\)

Answer 1

Correct option is (B) \( \frac {x^3}3 - x + \tan^{-1} x + c\)

\(\int \frac{x^{4}+1}{x^{2}+1} \mathrm{~d} x = \int \frac {x^4 - 1 + 1+1}{x^2 + 1}dx\)

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

\(= \int \frac{(x^2 + 1) (x^2 - 1)}{(x^2 +1)} dx + 2\int \frac 1{x^2 + 1} dx\)

\(= \int (x^2 - 1)dx + \int \frac 1{x^2 + 1} dx\)

\(= \int x^2dx - \int 1 dx+ \int \frac 1{x^2 + 1} dx\)

\(= \frac{x^3 }3 - x + \tan^{-1}x + c\)