int frac{x^3-7 x+6}{x^2+3 x} ,d x= | vedclass

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(C) First, perform polynomial long division on the integrand $\frac{x^3-7x+6}{x^2+3x}$.Dividing $x^3-7x+6$ by $x^2+3x$ gives a quotient of $(x-3)$ and

Vedclass · Accepted Answer

$\frac{x^2}{2}-3 x+2 \log |x|+c$, where $c$ is a constant of integration.

Vedclass · Answer

(C) First, perform polynomial long division on the integrand $\frac{x^3-7x+6}{x^2+3x}$.Dividing $x^3-7x+6$ by $x^2+3x$ gives a quotient of $(x-3)$ and a remainder of $(2x+6)$.Thus, $\frac{x^3-7x+6}{x^2+3x} = x-3 + \frac{2x+6}{x^2+3x}$.Simplify the remainder term: $\frac{2x+6}{x^2+3x} = \frac{2(x+3)}{x(x+3)} = \frac{2}{x}$ for $x 
eq -3$.Now, integrate the expression: $\int (x-3+\frac{2}{x}) dx$.$= \int x dx - \int 3 dx + \int \frac{2}{x} dx$.$= \frac{x^2}{2} - 3x + 2 \log |x| + c$.

$\int \frac{x^3-7 x+6}{x^2+3 x} \,d x=$

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