(a) If \(y = (2x + 3)^{7} + \frac{x + 1}{2x - 1}\), find the
(a) If \(y = (2x + 3)^{7} + \frac{x + 1}{2x - 1}\), find the value of \(\frac{\mathrm d y}{\mathrm d x}\) at x = -1.
(b) Using the substitution, \(u = x + 2\), evaluate \(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x\).
Explanation
(a) \(y = (2x + 3)^{7} + \frac{x - 1}{2x - 1}\)
\(\frac{\mathrm d y}{\mathrm d x} = 7(2x + 3)^{6}. 2 + \frac{(2x - 1).1 - (x - 1).2}{(2x - 1)^{2}}\)
= \(14(2x + 3)^{6} + \frac{2x - 1 - 2x + 2}{(2x - 1)^{2}}\)
= \(14(2x + 3)^{6} + \frac{1}{(2x - 1)^{2}}\)
At x = -1, \(\frac{\mathrm d y}{\mathrm d x} = 14(-1)^{6} + \frac{1}{(2(-1) - 1)^{2}}\)
= \(\frac{\mathrm d y}{\mathrm d x} = 14 + \frac{1}{9} = 14\frac{1}{9}\)
(b) \(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x\)
\( u = x + 2 \implies \mathrm d u = \mathrm d x\)
If x = 1, u = 1 + 2 = 3; x = 2, u = 2 + 2 = 4.
\(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x = \int_{3} ^{4} \frac{u - 3}{u^{4}}\)
= \(\int_{3} ^{4} (\frac{u}{u^{4}} - \frac{3}{u^{4}}) \mathrm d u\)
= \(\int_{3} ^{4} (u^{-3} - 3u^{-4}) \mathrm d u\)
= \([\frac{u^{-2}}{-2} + u^{-3}]_{3} ^{4}\)
= \([\frac{1}{u^{3}} - \frac{1}{2u^{2}}]_{3} ^{4}\)
= \((\frac{1}{4^{3}} - \frac{1}{2(4^{2})}) - (\frac{1}{3^{3}} - \frac{1}{2(3^{2})})\)
= \(\frac{1}{64} - \frac{1}{32} - \frac{1}{27} + \frac{1}{18}\)
= \(\frac{5}{1728}\).