How do you find the integral of $7 x^{2} ln (x) d x$ $7x^2ln(x) dx$ ?

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How do you find the integral of $7 x^{2} ln (x) d x$ $7x^2ln(x) dx$ ?

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Answer 1 · 2015-03-08T19:28:33

I would use Integretion by Parts:
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Answer 2 · 2015-03-08T19:31:46

How: Use integration by parts.
(Any integral of the form $a x^{n} ln x$ $ax^nlnx$ can be done by parts.)

Details:
$\int 7 x^{2} ln (x) d x = 7 \int x^{2} ln x d x$ $int7x^2ln(x)dx=7intx^2lnxdx$

Let $u = ln x$ $u=lnx$ and $d v = x^{2} d x$ $dv=x^2dx$

This makes $d u = \frac{1}{x} d x$ $du=1/xdx$ and $v = \int x^{2} d x = (\frac{1}{3}) x^{3}$ $v=intx^2dx=(1/3)x^3$ (We'll add the $+ C$ $+C$ later.)

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

$7 \int x^{2} ln x d x = 7 [ln x (\frac{1}{3}) x^{3} - \int (\frac{1}{3}) x^{3} \cdot \frac{1}{x} d x]$ $7intx^2lnxdx=7[lnx(1/3)x^3-int(1/3)x^3*1/xdx]$

$= 7 [\frac{1}{3} x^{3} ln x - \frac{1}{3} \int \frac{x^{3}}{x} d x] = 7 [\frac{1}{3} x^{3} ln x - \frac{1}{3} \int x^{2} d x]$ $=7[1/3x^3lnx-1/3intx^3/xdx]=7[1/3x^3lnx-1/3intx^2dx]$

$= 7 [\frac{1}{3} x^{3} ln x - \frac{1}{3} (\frac{1}{3} x^{3})] + C$ $=7[1/3x^3lnx-1/3(1/3x^3)]+C$

$= \frac{7}{3} x^{3} ln x - \frac{7}{9} x^{3} + C$ $=7/3x^3lnx-7/9x^3+C$

General
Notice that the solution can be made quite general.
For example: with 5 instead of 3 the problem becomes $\int x^{5} ln x d x$ $intx^5lnxdx$ which can be solved by the same reasoning to get ; $\frac{1}{6} x^{6} ln x - \frac{1}{36} x^{6} + C$ $1/6x^6lnx-1/36x^6+C$

Note: For even better understanding, check the answers by differentiating.

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How do you find the integral of $7 x^{2} ln (x) d x$ $7x^2ln(x) dx$ ?

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