How do you integrate $\int (3 x ln (2 x) d x}$ $int( 3x ln(2x) dx}$ ?

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Answer 1 · 2018-04-10T06:35:00

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

Here,

$I = \int (3 x) ln (2 x) d x$ $I=int(3x)ln(2x)dx$

$Using Integration by Parts$ $"Using "color(blue)"Integration by Parts"$

$\int (u \cdot v) d x = u \int v d x - \int (\frac{d u}{d x} \int v d x) d x$ $color(red)(int(u*v)dx=uintvdx-int((du)/(dx)intvdx)dx$

Let, $u = ln (2 x) and v = 3 x$ $u=ln(2x)and v=3x$ ,we get

$\frac{d u}{d x} = \frac{1}{2 x} \cdot 2 = \frac{1}{x} and \int v d x = \frac{3 x^{2}}{2}$ $(du)/(dx)=1/(2x)*2=1/x and intvdx=(3x^2)/2$

$\Rightarrow I = ln (2 x) \frac{3 x^{2}}{2} - \int (\frac{1}{x} \frac{3 x^{2}}{2}) d x$ $=>I=ln(2x)(3x^2)/2-int(1/x(3x^2)/2)dx$

$= \frac{3 x^{2}}{2} ln (2 x) - \frac{3}{2} \int x d x + c$ $=(3x^2)/2ln(2x)-3/2intxdx+c$

$= \frac{3 x^{2}}{2} ln (2 x) - \frac{3}{2} \cdot \frac{x^{2}}{2} + C$ $=(3x^2)/2ln(2x)-3/2*x^2/2+C$

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

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

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

How do you integrate int( 3x ln(2x) dx}∫(3xln(2x)dx}int( 3x ln(2x) dx}?