How do you use integration by parts to establish the reduction formula $int(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dx$ ?

Question

How do you use integration by parts to establish the reduction formula $int(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dx$ ?

1 Answer

Impact of this question

13281 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-08-18T13:51:28

Remember that Integration by Parts involves the following:

$intudv = uv - intvdu$

To accomplish this, we must choose a term for $u$ , and another for $dv$ . To figure out which terms would work best, we can use the ILATE method for Integration by Parts:

I nverse
L ogarithm
A lgebraic
T rig
E xponential

This order of priority can help you decide which term should be our $u$ , and which should be our $dv$ . Whichever term in our equation that's higher on this list should be our $u$ . For this particular equation, you can ignore the $n$ and treat the $lnx$ on its own, making $(lnx)^n$ our $u$ , and $dx$ our $dv$ .

Now we must differentiate our $u$ and integrate our $dv$ , giving us:

$d/dx(lnx)^n = n(lnx)^(n-1)(1/x)$ ,

and

$int dx = x$

With our original Integration by Parts formula, that gives us:

$int (lnx)^ndx = x(lnx)^n - nint(lnx)^(n-1)(1/x)x$

Since the $x$ and the $1/x$ cancel out, we end up with the final reduction formula answer:

$int(lnx)^ndx = x(lnx)^n - nint(lnx)^(n-1)dx$

Science

Math

Humanities

... and beyond

How do you use integration by parts to establish the reduction formula $int(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dx$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use integration by parts to establish the reduction formula int(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dxint(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dx ?

1 Answer

Related questions

Impact of this question

How do you use integration by parts to establish the reduction formula $int(ln(x))^n dx = x(ln(x))^n-nint(ln(x))^(n-1)dx$ ?