How do you integrate $\int x {ln x}^{3} d x$ $int xln x^3 dx$ using integration by parts?

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Answer 1 · 2016-12-12T20:36:49

$I = \frac{3 x^{2}}{4} [2 ln x - 1] + C$ $I=(3x^2)/4[2lnx-1]+C$

formula is: $\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$ $" "intu(dv)/(dx)dx=uv-intv(du)/(dx)dx$

the choice of $u$ $" "u" "$ & $v$ $" "v$ is vital.

if logs are involved then it is usual to take $u = ln f (x)$ $u=lnf(x)$ but try and simplify the expression using the laws of logs first if possible.

$\int x {ln x}^{3} d x = \int 3 x ln x d x$ $intxlnx^3dx=int3xlnxdx$

$u = ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x}$ $u=lnx=>(du)/(dx)=1/x$

$\frac{d v}{d x} = 3 x \Rightarrow v = \frac{3 x^{2}}{2}$ $(dv)/(dx)=3x=>v=(3x^2)/2$

$I = \frac{3 x^{2}}{2} ln x - \int \frac{3 x^{2}}{2} \times \frac{1}{x} d x$ $I=(3x^2)/2lnx-int(3x^2)/2xx1/xdx$

$I = \frac{3 x^{2}}{2} ln x - \int \frac{3 x}{2} d x$ $I=(3x^2)/2lnx-int(3x)/2dx$

$I = \frac{3 x^{2}}{2} ln x - \frac{3 x^{2}}{4} + C$ $I=(3x^2)/2lnx-(3x^2)/4+C$

$I = \frac{3 x^{2}}{4} [2 ln x - 1] + C$ $I=(3x^2)/4[2lnx-1]+C$

How do you integrate int xln x^3 dx ∫xlnx3dxint xln x^3 dx using integration by parts?