How do you integrate ${(\frac{ln x}{3})}^{3}$ $(ln x / 3)^3$ ?

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How do you integrate ${(\frac{ln x}{3})}^{3}$ $(ln x / 3)^3$ ?

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Answer 1 · 2016-04-20T16:40:55

$\int {(\frac{ln x}{3})}^{3} d x = \frac{x {ln}^{3} x}{27} - \frac{x {ln}^{2} x}{9} + \frac{2 x ln x}{9} - \frac{2 x}{9} + C$ $int(lnx/3)^3dx=(xln^3x)/27-(xln^2x)/9+(2xlnx)/9-(2x)/9+C$

Explanation:

Note that ${(\frac{ln x}{3})}^{3} = \frac{{ln}^{3} x}{27}$ $(lnx/3)^3=ln^3x/27$ . From this, we see that

$\int {(\frac{ln x}{3})}^{3} d x = \frac{1}{27} \int {ln}^{3} x d x$ $int(lnx/3)^3dx=1/27intln^3xdx$

Using integration by parts:

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

We let

$u = {ln}^{3} x \Rightarrow d u = \frac{3 {ln}^{2} x}{x} d x$ $u=ln^3x" "=>" "du=(3ln^2x)/xdx$

$d v = (1) d x \Rightarrow v = x$ $dv=(1)dx" "=>" "v=x$

This gives us:

$\frac{1}{27} \int {ln}^{3} x d x = \frac{1}{27} x {ln}^{3} x - \frac{1}{27} \int 3 {ln}^{2} x d x$ $1/27intln^3xdx=1/27xln^3x-1/27int3ln^2xdx$

$= \frac{1}{27} x {ln}^{3} x - \frac{1}{9} \int {ln}^{2} x d x$ $=1/27xln^3x-1/9intln^2xdx$

Integrate $\int {ln}^{2} x d x$ $intln^2xdx$ similarly (using by parts again):

$u = {ln}^{2} x \Rightarrow d u = \frac{2 ln x}{x} d x$ $u=ln^2x" "=>" "du=(2lnx)/xdx$

$d v = (1) d x \Rightarrow v = x$ $dv=(1)dx" "=>" "v=x$

Thus,

$\int {ln}^{2} x = x {ln}^{2} x - \int 2 ln x d x$ $intln^2x=xln^2x-int2lnxdx$

Combining this and multiplying by $- \frac{1}{9}$ $-1/9$ , we see that:

$\frac{1}{27} \int {ln}^{3} x d x = \frac{x {ln}^{3} x}{27} - \frac{x {ln}^{2} x}{9} + \frac{2}{9} \int ln x d x$ $1/27intln^3xdx=(xln^3x)/27-(xln^2x)/9+2/9intlnxdx$

Use integration by parts one last time:

$u = ln x \Rightarrow d u = \frac{1}{x} d x$ $u=lnx" "=>" "du=1/xdx$

$d v = (1) d x \Rightarrow v = x$ $dv=(1)dx" "=>" "v=x$

Thus,

$\int ln x d x = x ln x - \int d x = x ln x - x$ $intlnxdx=xlnx-intdx=xlnx-x$

Hence,

$\frac{1}{27} \int {ln}^{3} x d x = \frac{x {ln}^{3} x}{27} - \frac{x {ln}^{2} x}{9} + \frac{2 x ln x}{9} - \frac{2 x}{9} + C$ $1/27intln^3xdx=(xln^3x)/27-(xln^2x)/9+(2xlnx)/9-(2x)/9+C$

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How do you integrate ${(\frac{ln x}{3})}^{3}$ $(ln x / 3)^3$ ?

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