Practicing Success
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The integral $∫\frac{x^4+2}{x^2+1}dx $ is equal to |
$\frac{1}{3}x^3-x+3tan^{-1}x+c,$ where c is a constant $\frac{1}{3}x^3+x+3tan^{-1}x+c,$ where c is a constant $\frac{1}{3}x^3+x-3tan^{-1}x+c,$ where c is a constant $\frac{1}{3}x^3-x-3tan^{-1}x+c,$ where c is a constant |
$\frac{1}{3}x^3-x+3tan^{-1}x+c,$ where c is a constant |
The correct answer is Option (1) → $\frac{1}{3}x^3-x+3tan^{-1}x+c,$ where c is a constant $∫\frac{x^4+2}{x^2+1}dx$ so $∫\frac{x^4+2}{x^2+1}dx=\int x^2-1+\frac{3}{x^2+1}dx$ $=\frac{x^3}{3}-x+3\tan^{-1}x+C$ using long division |
