Mathematics

Mathematics

Problem

Evaluate ∫3x3+6x2+x−2x2+2x+5 dx\displaystyle \int \dfrac{3x^{3}+6x^{2}+x-2}{x^{2}+2x+5}\,dx. a) 32x2−7ln⁡∣x2+2x+5∣+3arctan⁡ ⁣(x+12)+C\dfrac{3}{2}x^{2}-7\ln|x^{2}+2x+5|+3\arctan\!\left(\dfrac{x+1}{2}\right)+C b) x2−7ln⁡∣x2+2x+5∣+6arctan⁡ ⁣(x+12)+Cx^{2}-7\ln|x^{2}+2x+5|+6\arctan\!\left(\dfrac{x+1}{2}\right)+C c) 32x2−7ln⁡∣x2+2x+5∣+6arctan⁡ ⁣(x+12)+C\dfrac{3}{2}x^{2}-7\ln|x^{2}+2x+5|+6\arctan\!\left(\dfrac{x+1}{2}\right)+C d) 32x2+7ln⁡∣x2+2x+5∣+6arctan⁡ ⁣(x+12)+C\dfrac{3}{2}x^{2}+7\ln|x^{2}+2x+5|+6\arctan\!\left(\dfrac{x+1}{2}\right)+C
Intuitive explanation

The key is polynomial long division: since the numerator has degree 3 and the denominator has degree 2, you can divide to extract a linear quotient plus a remainder. This splits the integral into an easy polynomial part and a rational part where the numerator is now lower degree than the denominator. For that remainder fraction, complete the square in the denominator to unlock a standard arctangent form.

Intuitive explanation

The key is polynomial long division: since the numerator has degree 3 and the denominator has degree 2, you can divide to extract a linear quotient plus a remainder. This splits the integral into an easy polynomial part and a rational part where the numerator is now lower degree than the denominator. For that remainder fraction, complete the square in the denominator to unlock a standard arctangent form.

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