Reduction Formula

[sareen]

Use integration by parts to prove the reduction formula:
?(lnx)^n dx= x(lnx)^n - n?(lnx)^n-1 dx

Then use the above to evaluate the integral:
? (lnx)^3 dx

 

[soroban]

Hello, sareen!

Use integration by parts to prove the reduction formula:
. . $\displaystyle \int(\ln x)^n\,dx \;=\; x(\ln x)^n - n\int(\ln x)^{n-1}dx$

$\displaystyle \text{Let: }\;\begin{array}{ccccccc}u &=& (\ln x)^n & & dv &=& dx \\ du &=& n(\ln x)^{n-1}\frac{dx}{x} && v &=& x \end{array}$


$\displaystyle \text{Then: }\;\int\underbrace{(\ln x)^x}_u\,\underbrace{dx}_{dv} \;=\;\underbrace{(\ln x)^n}_u\cdot \underbrace{x}_v - \int\underbrace{x}_v\cdot\underbrace{ n(\ln x)^{n-1}\tfrac{dx}{x}}_{du}$

. . . . . . . . . . . . . . . .$\displaystyle =\; x(\ln x)^n - n\int (\ln x)^{n-1}dx$

Then use the above to evaluate the integral:
. . $\displaystyle \int (\ln x)^3 dx$

$\displaystyle \int(\ln x)^3dx \;=\; x(\ln)^3 - 3\underbrace{\int(\ln x)^2dx}$
. . . . . . . . $\displaystyle = \; x(\ln x)^3 - 3\overbrace{ \bigg[x(\ln x)^2 - 2\!\!\int \ln x\,dx\bigg]}^{\searrow\quad} + C$

. . . . . . . . $\displaystyle =\; x(\ln x)^3 - 3x(\ln x)^2 + 6\underbrace{\int \ln x\,dx}_{\quad\searrow} + C$
. . . . . . . . $\displaystyle =\; x(\ln x)^3 - 3x(\ln x)^2 + 6\overbrace{\bigg[x\ln x - \int dx\bigg]} + C$

. . . . . . . . $\displaystyle =\; x(\ln x)^3 - 3x(\ln x)^2 + 6\bigg[x\ln x - x\bigg] + C$

. . . . . . . . $\displaystyle =\; x(\ln x)^3 - 3x(\ln x)^2 + 6x\ln x - 6x + C$

 

[sareen]

thanks a lot! It really cleared out all of my questions