How do you integrate $\int 4 x^{2} {ln x}^{2} d x$ $int 4 x^2 ln x^2 dx$ using integration by parts?

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How do you integrate $\int 4 x^{2} {ln x}^{2} d x$ $int 4 x^2 ln x^2 dx$ using integration by parts?

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Answer 1 · 2016-04-19T21:58:12

$\int 4 x^{2} ln (x^{2}) d x = \frac{8 x^{3} (3 ln x - 1)}{9} + C$ $int4x^2ln(x^2)dx=(8x^3(3lnx-1))/9+C$

Explanation:

First, rewrite $ln (x^{2}) = 2 ln x$ $ln(x^2)=2lnx$ using the rule that $ln (a^{b}) = b \cdot ln a$ $ln(a^b)=b*lna$ .

This gives us the integral $\int 4 x^{2} (2 ln x) d x = 8 \int x^{2} ln x d x$ $int4x^2(2lnx)dx=8intx^2lnxdx$ .

For this, recalling that integration by parts takes the form $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ , we let

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

$d v = x^{2} d x \Rightarrow \int d v = \int x^{2} d x \Rightarrow v = \frac{x^{3}}{3}$ $dv=x^2dx" "=>" "intdv=intx^2dx" "=>" "v=x^3/3$

This gives us:

$8 \int x^{2} ln x = 8 [(\frac{x^{3}}{3}) ln x - \int \frac{x^{3}}{3} (\frac{1}{x}) d x]$ $8intx^2lnx=8[(x^3/3)lnx-intx^3/3(1/x)dx]$

$= \frac{8 x^{3} ln x}{3} - 8 \int \frac{x^{2}}{3} d x$ $=(8x^3lnx)/3-8intx^2/3dx$

$= \frac{8 x^{3} ln x}{3} - \frac{8 x^{3}}{9} + C$ $=(8x^3lnx)/3-(8x^3)/9+C$

Which can also be written as

$\int 4 x^{2} ln (x^{2}) d x = \frac{8 x^{3} (3 ln x - 1)}{9} + C$ $int4x^2ln(x^2)dx=(8x^3(3lnx-1))/9+C$

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How do you integrate $\int 4 x^{2} {ln x}^{2} d x$ $int 4 x^2 ln x^2 dx$ using integration by parts?

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How do you integrate int 4 x^2 ln x^2 dx ∫4x2lnx2dxint 4 x^2 ln x^2 dx using integration by parts?

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How do you integrate $\int 4 x^{2} {ln x}^{2} d x$ $int 4 x^2 ln x^2 dx$ using integration by parts?