MHTCET 16 May 2024 Shift 2 Paper

Concept:We simplify the integrand using polynomial division and then cancel common factors before integrating term by term.Explanation:The given integral is $I = \int \frac{x^{3}-7x+6}{x^{2}+3x} \, dx$.First, divide $x^{3}-7x+6$ by $x^{2}+3x$ because the degree of the numerator is higher.The division gives a quotient $x-3$ and remainder $2x+6$.So, $\frac{x^{3}-7x+6}{x^{2}+3x} = x-3 + \frac{2x+6}{x^{2}+3x}$.Now factor the denominator: $x^{2}+3x = x(x+3)$.Also, $2x+6 = 2(x+3)$.Thus, $\frac{2x+6}{x^{2}+3x} = \frac{2(x+3)}{x(x+3)} = \frac{2}{x}$, for $x \neq -3, 0$.So the integral becomes $I = \int \left(x - 3 + \frac{2}{x}\right) dx$.Integrating each term, we get $\int x \, dx = \frac{x^{2}}{2}$, $\int (-3) \, dx = -3x$, and $\int \frac{2}{x} \, dx = 2\ln|x|$.Therefore, $I = \frac{x^{2}}{2} - 3x + 2\ln|x| + c$, where $c$ is the constant of integration.Answer:$\frac{x^{2}}{2} - 3x + 2 \ln|x| + c$Hence, the correct option is C.