Posted by **Jules** on Monday, April 25, 2011 at 1:55pm.

Could someone work this question out so I understand it.

Find the indefinite intregral

(lnx)^7 / x dx

Use C as the arbitrary constant.

- Math -
**Anonymous**, Monday, April 25, 2011 at 2:02pm
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

- Math -
**MathMate**, Monday, April 25, 2011 at 2:36pm
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

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