Question

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

• A. 13
• B. 11
• C. 5
• D. 8

Answer 1

Answer: (C)

$\int x^3(\ln x{)}^{2}dx=\frac{x^4}{32}\left(a(\ln x{)}^2+b(\ln x)+c\right)+d$
Differentiating both sides,
$x^3(\ln x{)}^2=\frac{1}{32}{x}^4\left(\frac{a\cdot 2\ln x}{x}+\frac{b}{x}\right)+\frac{4x^3}{32}\left(a(\ln x{)}^2+b(\ln x)+c\right)$
$x^3(\ln x{)}^2=\frac{a}{16}{x}^3\ln x+\frac{b}{32}{x}^3+\frac{a}{8}{x}^3(\ln x{)}^2+\frac{b}{8}{x}^3(\ln x)+\frac{c{x}^3}{8}$
$x^3(\ln {)}^2=\left(\frac{a}{16}+\frac{b}{8}\right)x^3\ln x+\frac{a}{8}{x}^3(\ln x{)}^2+\left(\frac{b}{32}+\frac{c}{8}\right)x^3$
Comparing both sides,
$\frac{a}{16}+\frac{b}{8}=0;\frac{a}{8}=1\text{ and }\frac{b}{32}+\frac{1}{8}=0$
$$\begin{gathered} a+2b=0\Rightarrow a=8b+4c=0\Rightarrow c=1 \\ <br/>b=-4 \end{gathered}$$
$\therefore a+b+c=8-4+1=5$