This particular integral is driving me crazy. I can't figure it out for the life of me. Could someone please explain to me how to go about solving it?
Integrate[5x(7x - 9)^14dx
I was given a few hints but they still are not helping me at all. First let U=7x-9 and then let x=(U+9)/7. I would imagine this should be done using the substitution method. Thanks in advance
John
Is thatOriginally Posted by TreeMoney
INT.[5x (7x -9)^14]dx ?
If yes, then here is one way of doing it.
Integration by parts:
INT.[u]dv = u*v -INT.[v]du ------***
Let u = 5x
So, du = 5dx
And dv = (7x -9)^14 dx = (7x-9)^14 *(7dx /7) = (1/7)(7x-9)^14 *(7dx)
So, v = (1/7)[(1/15)(7x-9)^15] = (1/105)(7x-9)^15
Hence,
INT.[5x (7x-9)^14]dx
= (5x)(1/105)(7x-9)^15 -INT.[(1/105)(7x-9)^15](5dx)
= (1/21)x(7x-9)^15 -(1/21)INT[(7x-9)^15]dx
= (1/21)x(7x-9)^15 -(1/21){(1/7)INT[(7x-9)^15](7dx)}
= (1/21)x(7x-9)^15 -(1/147){(1/16)(7x-9)^16}
= (1/21)x(7x-9)^15 -(1/2352)(7x-9)^16 -------------answer.
Check,
d/dx {(1/21)x(7x-9)^15 -(1/2352)(7x-9)^16}
= (1/21){[x*15(7x-9)^14 *7] +[(7x-9)^15 *1]} -(1/2352){16(7x-9)^15 *7}
= {5x(7x-9)^14 +(1/21)(7x-9)^15} -(1/21)(7x-9)^15
= 5x(7x-9)^14
Okay.
It can be done by integration by parts, but also directly by substitution.
$\displaystyle \int dx \, 5x(7x-9)^{14} = \int \left ( \frac{du}{7} \right ) \, 5 \left ( \frac{u+9}{7} \right ) u^{14}$
$\displaystyle =\frac{5}{7} \int du \, \left ( \frac{u}{7} + \frac{9}{7} \right )u^{14}$
$\displaystyle =\frac{5}{7} \int du \, \frac{1}{7}u^{15} + \frac{5}{7} \int du \, \frac{9}{7}u^{14}$
$\displaystyle =\frac{5}{7} \frac{1}{7}\frac{1}{16}u^{16} + \frac{5}{7} \frac{9}{7}\frac{1}{15}u^{15}$
$\displaystyle =\frac{5}{784}(7x-9)^{16}+\frac{3}{49}(7x-9)^{15}$
Which you can then simplify any way you wish.
-Dan
There's no such thing as a "stupid mistake." You either get it right or you don't. You didn't get it right, you knew you didn't, and you asked a question about it. I see no stupidity there, I see a good practice in action!
-Dan
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