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∫ 9 (3𝑥 + 10) 6

 Jan 12, 2021
 #1
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Is this your integral?

 

\(\displaystyle\int9(3x+10)^6 dx\)

 

If so....

 

Let's pick   u = 3x + 10   which means  du = 3 dx   and then substitute these in like this:

 

\(\phantom{=\quad}\displaystyle\int9(3x+10)^6 dx\\~\\ {=\quad}\displaystyle\int3\cdot3(3x+10)^6 dx\\~\\ {=\quad}3\displaystyle\int(3x+10)^6 \cdot 3dx\\~\\ {=\quad}3\displaystyle\int u^6 \cdot du \\~\\ {=\quad}3\displaystyle\int u^6 du\)

                                           Now we can use the rule which says  \(\int u^n du\ =\ \frac{u^{n+1}}{n+1}+c\)    to get...

\({=\quad}3\cdot\frac{u^7}{7}+c\\~\\ {=\quad}\frac37u^7+c\)

                                           Now let's substitute  3x + 10  back in for  u

\({=\quad}\frac37(3x+10)^7+c\)

 Jan 12, 2021

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