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a) $\displaystyle \int_{-1}^0 x \sqrt{2 - 5x}dx. $

I think I should use some change of variables here, but it didn't work the way I tried to use it.

b) $\displaystyle \int_{0}^{2} \frac{1}{(x + 2)(x + 3)}dx.$

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$\frac{1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}$, partial fraction decomposition –  Cortizol Feb 20 '13 at 19:25
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a) Substitute $u=2-5x$. b) Partial fractions. –  David Mitra Feb 20 '13 at 19:26

2 Answers 2

up vote 1 down vote accepted

Per Mitra:

$u=2-5x \iff du=-5\,dx$,

$$\therefore \int_{-1}^{0} x\sqrt{2-5x} \, dx=\int_{-1}^{0}\frac{u-2}{-5}\sqrt{u}\frac{1}{-5} \, du.$$

$$ \frac{1}{(x+2)(x+3)}=\frac{1}{x+2}+\frac{-1}{x+3}. $$

$$\therefore \int_0^2 \frac{1}{(x+2)(x+3)} \, dx=\int_0^2\frac{1}{x+2} \, dx-\int_0^2 \frac{1}{x+3}.$$

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Thanks a lot! So we have $log(6/5)$ in the second one.. –  Michael Feb 20 '13 at 19:37
    
@Michael, indeed. Have a good day. –  000 Feb 20 '13 at 19:39

Another way is by a rationalizing substitution: $$ \begin{align} u & = \sqrt{2-5x} \\[8pt] u^2 & = 2 - 5x \\[8pt] 2u\,du & = -5\,dx \\[8pt] \frac{-2}{5}\, du & = dx \\[8pt] x & = \frac{2-u^2}{5} \end{align} $$ When $x=-1$ then $u=\sqrt{7}$ and when $x=0$ then $u=\sqrt{2}$. So $$ \int_{-1}^0 x \sqrt{2-5x} \, dx = \int_\sqrt{7}^\sqrt{2} \frac{2-u^2}{5}\cdot u\left(\frac{-2}{5}\, du \right) = \frac{4}{25}\int_\sqrt{2}^\sqrt{7}(2u-u^3)\,du $$ $$ = \frac{4}{25} \left[u^2-\frac{u^4}{4}\right]_\sqrt{2}^\sqrt{7} = \frac{4}{25}\left( \left(7-\frac{49}{4}\right) - (2-4) \right) =\cdots $$

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