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Integration by parts- what is $\int u’v’\mathrm dx$?
This is what I got:
$u = x^2 + \ln x$
$u’ = 2x + 1/x$
$v = x^3$
$v’ = 3x^2$
Hence:
$\int u’v’ = \int(2x + 1/x)(3x^2) = \int(8x^3 + 9/x^2) = \frac{8}{3}x^3 + 9 \ln(x) + C$
But the textbook’s answer is: $\frac{8}{3}x^3 + 9 \ln(x)$
Where is my mistake?
A:
The typical trick used in such situations is to take the product rule:
$$
u’v’ = u v’ + u’v
$$
So
$$
u’v’ = u v’ + u’v \\
\Rightarrow \frac{8}{3}x^3 + \frac{9}{x} + 9 \ln x + C = \frac{8}{3}x^3 + 9 \ln x + 9 \ln x + C = \frac{8}{3}x^3 + 9 \ln x \\
\Rightarrow \frac{8}{3}x^3 + 9 \ln x = \frac{8}{3}x^3 + 9 \ln x + 9 \ln x + C
$$
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