confused and rusty on this any help appreciated (1 Viewer)

ekjchale#1 · May 25, 2022

if i do du/dy it is1 and idk wut to do after, ive done this b4 but have forgotten so any assistance is appreciated

Drongoski · May 25, 2022

ekjchale#1 said:
View attachment 35659if i do du/dy it is1 and idk wut to do after, ive done this b4 but have forgotten so any assistance is appreciated

a)
$\int y\sqrt{y+1}]dy = \int (y+1 - 1)\sqrt {y+1} d(y+1) \\ \\ = \int ((y+1)^{\frac {3}{2}} - (y+1)^\frac{1}{2}) d(y+1) = \frac{2}{5}(y+1)^\frac{5}{2} - \frac{2}{3}(y+1)^\frac {3}{2} + C$

c)
$\int \frac {x}{\sqrt{2x-1}} dx = \frac {1}{2}\times \frac {1}{2}\int \frac {(2x-1 + 1)}{\sqrt{2x-1}} d(2x-1) \\ \\ = \frac {1}{4}\int ((2x-1)^\frac {1}{2} + (2x-1)^\frac{-1}{2}) d(2x - 1)\\ \\ = \frac {1}{4}[( \frac {2}{3})(2x-1)^\frac {3}{2} + 2(2x-1)^\frac {1}{2}]+ C$

Note:
d(y+1) = dy
dx = 0.5d(2x-1)

I'm not doing it the conventional way; I skip the formal substitution. Most of you are not comfortable using differentials like d(2x-1).

b) By substitution

Let u = x-1 Therefore du/dx = 1 so that du = dx and x = u+1

$\therefore I = \int \frac {u + 1}{u^3} du = \int (u^{-2} + u^{-3})du = -\frac {1}{u} -\frac {1}{2u^2} + C \\ \\ = \frac {-1}{x-1} - \frac {1}{2(x-1)^2} + C$

confused and rusty on this any help appreciated (1 Viewer)

ekjchale#1

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Drongoski

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