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luda_lava [24]
4 years ago
13

\ (x^3-6x^2+9x+3)(3x^2-12x+9) dx" alt="\int\ (x^3-6x^2+9x+3)(3x^2-12x+9) dx" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
AlexFokin [52]4 years ago
6 0

If the integral is simply

\displaystyle\int(x^3-6x^2+9x+3)(3x^2-12x+9)\,\mathrm dx

then notice that

\mathrm d(x^3-6x^2+9x+3)=(3x^2-12x+9)\,\mathrm dx

which means you can compute the integral easily with a substitution

u=x^3-6x^2+9x+3\implies\mathrm du=(3x^2-12x+9)\,\mathrm dx

Under this transformation, the integral is

\displaystyle\int u\,\mathrm du=\frac{u^2}2+C=\boxed{\frac{(x^3-6x^2+9x+3)^2}2+C}

On the other hand, in case you're missing a symbol and the integral is actually

\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx

then first carry out the division:

\dfrac{x^3-6x^2+9x+3}{3x^2-12x+9}=\dfrac x3-\dfrac23-\dfrac{2x-9}{3x^2-12x+9}

Now, 3x^2-12x+9=3(x-3)(x-1), so to integrate the remainder term you can decompose it into partial fractions:

-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac a{x-3}+\dfrac b{x-1}

9-2x=a(x-1)+b(x-3)

x=1\implies7=-2b\implies b=-\dfrac72

x=3\implies3=2a\implies a=\dfrac32

\implies-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac 3{2(x-3)}-\dfrac 7{2(x-1)}

Then the integral would be

\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx=\boxed{\frac{x^2}6-\frac{2x}3+\frac32\ln|x-3|-\frac72\ln|x-1|+C}

which can be rewritten in several ways, such as

\dfrac{x^2-4x}6+\dfrac12ln\left|\dfrac{(x-3)^3}{(x-1)^7}\right|+C

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