Given that $\frac{6+5x-8x^2}{(2x-1)(x^2+3)} = \frac{A}{2x-1} + \frac{Bx+C}{x^2+3}$, where $A$, \

Question

Given that $\frac{6+5x-8x^2}{(2x-1)(x^2+3)} = \frac{A}{2x-1} + \frac{Bx+C}{x^2+3}$, where $A$, $B$ and $C$ are constants find the values of $A$, $B$ and $C$, differentiate $\ln(x^2 + 3)$ with respect to $x$. Using the results from parts (a) and (b), find $\displaystyle\int \frac{6+5x-8x^2}{(2x-1)(x^2+3)} \, \mathrm{d}x$.

Accepted Answer

(a) $A=2, B=-5, C=0$. (b) $\frac{2x}{x^2+3}$. (c) $\ln|2x-1|-\frac52\ln(x^2+3)+C$.

Techniques of Differentiation - Derivatives of Natural Logarithmic Functions

Techniques of Differentiation - Derivatives of Natural Logarithmic Functions

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