Evaluation of $$\int x^{26}(x-1)^{17}(5x-3) \, dx$$
I did not understand what substution i have used so that it can simplify,
I have seems it is a derivative of some function.
Help me, Thanks
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Note that \begin{align} \frac{d}{dx}\left[\color{blue}{\frac{1}{9}x^{27}(x-1)^{18}}\right]&=2x^{27}(x-1)^{17}+3x^{26}(x-1)^{18}\\ &=x^{26}(x-1)^{17}(2x+3(x-1))\\ &=x^{26}(x-1)^{17}(5x-3). \end{align} Intuition: given the form of the integrand, I played around with $Cx^{27}(x-1)^{18}$ and found $C=\frac{1}{9}$ worked.
yurnero
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You have this: $$\left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)$$
Perhaps the only simple way to do this is to recall what you see above is what you get when you evaluate $$ \frac d {dx } \left( \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \right) $$ For example, \begin{align} & \frac d {dx} (4x+19)^{42} (2x-27)^{50} \\[10pt] = {} & \underbrace{\overbrace{42(4x+19)^{41}\cdot4\cdot (2x-27)^{50}} {}+{} \overbrace{(4x+19)^{42} 50(2x-27)^{49}\cdot 2}}_\text{product rule} \\[10pt] = {} & \Big( \underbrace{(4x+19)^{41} (2x-27)^{49}}_\text{the common factor} \Big) \cdot \Big( \text{whatever is left (a sum of two terms, admitting simplification)} \Big) \\[10pt] = {} & \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left( \underbrace{ \begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} }_\text{This is “whatever is left.''} \right) \end{align}
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Even if the exponent is not "large", this will still be true... – MasB Nov 13 '16 at 16:39
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@BernardMasse : True, but if it's not "large", you might just expand the whole expression before integrating. $\qquad$ – Michael Hardy Nov 13 '16 at 16:40
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It depends what you call large. How about 5 and 3? Anyways, I found your answer interesting because you used a very helpful way of writing it, including the "whatever is left" – MasB Nov 13 '16 at 16:45
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@BernardMasse : I'm glad you liked it. – Michael Hardy Nov 13 '16 at 17:05
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Sorry for the informal writing am rushing out.
Hope it helps! $$ (x - 1)^{17} = \sum \left( \begin{array}{c} 17 \\ r \end{array} \right) x^{17-r}(-1)^r \\ x^{26}(x - 1)^{17} = \sum \left( \begin{array}{c} 17 \\ r \end{array} \right) x^{43-r}(-1)^r \\ x^{26}(x - 1)^{17}(5x−3) = 5\sum \left( \begin{array}{c} 17 \\ r \end{array} \right) x^{45-r}(-1)^r\space- 3\sum\left( \begin{array}{c} 17 \\ r \end{array} \right) x^{44-r}(-1)^r \\ =5x^{45} - \left(\sum\left[ 5\left( \begin{array}{c} 17 \\ r+1 \end{array}\right) - 3\left( \begin{array}{c} 17 \\ r \end{array}\right) \right]x^{44-r}(-1)^r\right) +3x^{27} \\ \int x^{26}(x - 1)^{17}(5x−3) = 5\int x^{45} - \left(\sum\left[ 5\left( \begin{array}{c} 17 \\ r+1 \end{array}\right) - 3\left( \begin{array}{c} 17 \\ r \end{array}\right) \right]\int x^{44-r}(-1)^r\right) \\+3\int x^{27} + c $$
Toye Brainz
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