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Q.
If $\int x ^3(\ln x )^2 dx =\frac{ x ^4}{32}\left( a (\ln x )^2+ b (\ln x )+ c \right)+ d$, where ' $d$ ' is the constant of integration, then $( a + b + c )$ is equal to
Integrals
Solution:
$\int x ^3(\ln x )^2 dx =\frac{ x ^4}{32}\left( a (\ln x )^2+ b (\ln x )+ c \right)+ d$
Differentiating both sides,
$x ^3(\ln x )^2=\frac{1}{32} x ^4\left(\frac{ a \cdot 2 \ln x }{ x }+\frac{ b }{ x }\right)+\frac{4 x ^3}{32}\left( a (\ln x )^2+ b (\ln x )+ c \right) $
$x ^3(\ln x )^2=\frac{ a }{16} x ^3 \ln x +\frac{ b }{32} x ^3+\frac{ a }{8} x ^3(\ln x )^2+\frac{ b }{8} x ^3(\ln x )+\frac{ cx ^3}{8}$
$x ^3(\ln )^2=\left(\frac{ a }{16}+\frac{ b }{8}\right) x ^3 \ln x +\frac{ a }{8} x ^3(\ln x )^2+\left(\frac{ b }{32}+\frac{ c }{8}\right) x ^3$
Comparing both sides,
$\frac{ a }{16}+\frac{ b }{8}=0 ; \frac{ a }{8}=1 \text { and } \frac{ b }{32}+\frac{1}{8}=0 $
$a +2 b =0 \Rightarrow a =8 b +4 c =0 \Rightarrow c =1 \\
b =-4$
$\therefore a + b + c =8-4+1=5$