If $\int x^{3} (ln x)^{2} d x = \frac{x^{4}}{32} (a (ln x)^{2} + b (ln x) + c) + d$ , where $d$ is the constant of integration, then $(a + b + c)$ is equal to

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Solution

$\int x^{3} (ln x)^{2} d x = \frac{x^{4}}{32} (a (ln x)^{2} + b (ln x) + c) + d$
Differentiating both sides w.r.t. $x$ , we get
$x^{3} (ln x)^{2} = \frac{x^{3}}{8} (a (ln x)^{2} + b (ln x) + c) + \frac{x^{4}}{32} (\frac{2 a ln x}{x} + \frac{b}{x})$
$\Rightarrow x^{3} (ln x)^{2} = (\frac{a}{16} + \frac{b}{8}) x^{3} ln x + \frac{a}{8} x^{3} (ln x)^{2} + (\frac{b}{32} + \frac{c}{8}) x^{3}$
Comparing both sides, we get
$\frac{a}{8} = 1 \Rightarrow a = 8$
$\frac{a}{16} + \frac{b}{8} = 0 \Rightarrow b = - 4$
$\frac{b}{32} + \frac{c}{8} \Rightarrow c = 1$
$∴ a + b + c = 8 - 4 + 1 = 5$

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