3.4c Integration as the Inverse of Differentiation

3.4c Integration as the Inverse of Differentiation

Example:

$\begin{array}{l} Shows that \frac{d}{d x} [\frac{2 x + 5}{x^{2} - 3}] = \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} \\ Hence, find the value of \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x \end{array}$

Solution:

$\begin{array}{l} \frac{d}{d x} [\frac{2 x + 5}{x^{2} - 3}] = \frac{(x^{2} - 3) (2) - (2 x + 5) (2 x)}{{(x^{2} - 3)}^{2}} \\ = \frac{2 x^{2} - 6 - 4 x^{2} - 10 x}{{(x^{2} - 3)}^{2}} \\ = \frac{- 2 x^{2} - 10 x - 6}{{(x^{2} - 3)}^{2}} \\ = \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} \\ \int_{0}^{2} \frac{- 2 (x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = {[\frac{2 x + 5}{x^{2} - 3}]}_{0}^{2} \\ - 2 \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = {[\frac{2 x + 5}{x^{2} - 3}]}_{0}^{2} \\ \int_{0}^{2} \frac{(x^{2} + 5 x + 3)}{{(x^{2} - 3)}^{2}} d x = - \frac{1}{2} [(\frac{2 (2) + 5}{2^{2} - 3}) - (\frac{2 (0) + 5}{0^{2} - 3})] \\ = - \frac{1}{2} [9 - (- \frac{5}{3})] \\ = - \frac{1}{2} \times \frac{32}{3} \\ = - \frac{16}{3} \\ = - 5 \frac{1}{3} \end{array}$

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