Asked by Jules
Could someone work this question out so I understand it.
Find the indefinite intregral
(lnx)^7 / x dx
Use C as the arbitrary constant.
Find the indefinite intregral
(lnx)^7 / x dx
Use C as the arbitrary constant.
Answers
Answered by
Anonymous
The best method here is integration by parts. Most of the times when you see products of functions AND the product of one with the others derivative gives a simple to integrate function.
Do you see that [ln(x)]' = 1/x. And when you multiply it with a power of x you get a power of x?
The standard form of the integration by parts is:
integral_{f'(x)*g(x)}dx = f(x)*g(x) - integral_{f(x)*g'(x)}dx
The idea is to pick f and g so it'll all be simpler. We already picked out g(x) = ln(x)
Now we need f'(x). This can only be x^7. But what functions derivative is x^7? We have to to the opposite of derivation ---> integration!
f(x) = integral_{f'(x)}dx = integral_{x^7}dx = (x^8)/8.
Just check! [(x^8)/8]' = [(1/8)*(x^8)]' = (1/8)*8*(x^7) = x^7
integral_{f'(x)*g(x)}dx = integral_{(x^7)*ln(x)}dx = integral_{ [(x^8)/8]' * ln(x) }dx =
= [(x^8)/8] * ln(x) - integral_{ [(x^8)/8] * ln'(x) }dx =
= [(x^8)/8] * ln(x) - integral_{ [(1/8)*(x^8)] * 1/x }dx =
= [(x^8)/8] * ln(x) - (1/8)* integral_{ [x^(8-1)] }dx =
= [(x^8)/8] * ln(x) - (1/8)* integral_{ [x^7] }dx =
= [(x^8)/8] * ln(x) - (1/8)*(x^8)/8 + real_constant =
= [(x^8)/8] * ln(x) - (x^8)/64 + real_constant = [(x^8)/8] * [ln(x) - (1/8)] + real_constant
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . = [(x^8)/64] * [ 8*ln(x) - 1 ] + real_constant
Do you see that [ln(x)]' = 1/x. And when you multiply it with a power of x you get a power of x?
The standard form of the integration by parts is:
integral_{f'(x)*g(x)}dx = f(x)*g(x) - integral_{f(x)*g'(x)}dx
The idea is to pick f and g so it'll all be simpler. We already picked out g(x) = ln(x)
Now we need f'(x). This can only be x^7. But what functions derivative is x^7? We have to to the opposite of derivation ---> integration!
f(x) = integral_{f'(x)}dx = integral_{x^7}dx = (x^8)/8.
Just check! [(x^8)/8]' = [(1/8)*(x^8)]' = (1/8)*8*(x^7) = x^7
integral_{f'(x)*g(x)}dx = integral_{(x^7)*ln(x)}dx = integral_{ [(x^8)/8]' * ln(x) }dx =
= [(x^8)/8] * ln(x) - integral_{ [(x^8)/8] * ln'(x) }dx =
= [(x^8)/8] * ln(x) - integral_{ [(1/8)*(x^8)] * 1/x }dx =
= [(x^8)/8] * ln(x) - (1/8)* integral_{ [x^(8-1)] }dx =
= [(x^8)/8] * ln(x) - (1/8)* integral_{ [x^7] }dx =
= [(x^8)/8] * ln(x) - (1/8)*(x^8)/8 + real_constant =
= [(x^8)/8] * ln(x) - (x^8)/64 + real_constant = [(x^8)/8] * [ln(x) - (1/8)] + real_constant
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . = [(x^8)/64] * [ 8*ln(x) - 1 ] + real_constant
Answered by
MathMate
Sometimes there are different ways to solve a problem, here is another:
∫(lnx)^7 / x dx
We note that (lnx)' = 1/x
and substitute y=ln(x), dy=dx/x
∫(lnx)^7 / x dx
= ∫ y^7 dy
= y^8/8 + C
= ln(x)^8/8 + C
∫(lnx)^7 / x dx
We note that (lnx)' = 1/x
and substitute y=ln(x), dy=dx/x
∫(lnx)^7 / x dx
= ∫ y^7 dy
= y^8/8 + C
= ln(x)^8/8 + C