For $x>0$, if $\int(\log x)^5 d x$ is equal to $x\left[A(\log x)^5+\right.$ $\left.B(\log x)^4+C(\log…
For $x>0$, if $\int(\log x)^5 d x$ is equal to $x\left[A(\log x)^5+\right.$ $\left.B(\log x)^4+C(\log x)^3+D(\log x)^2+E(\log x)+F\right]+$ constant, then $A+B+C+D+E+F$ is equal to
$-44$
$-42$
$-40$
$-36$
Solution
Given that,
$\begin{aligned}
\int\left(\log x^2\right) d x & =x\left[A(\log x)^5+B(\log x)^4\right] \\
& \left.+C(\log x)^3+D(\log x)^2+E(\log x)+F\right]+C
\end{aligned}$
Let $I=\int(\log x)^5 d x$
Put $\log x=t \quad \Rightarrow x=e^t \Rightarrow \mathrm{d} x=e^t d t$
$I=\int t^5 e^5 d t$
$I=e^t\left[t^5-5 t^4+20 t^3-60 t^2+120 t-120\right]+C$
Now, put $\mathrm{E}=\log x$
$\begin{aligned}
& I=x\left[(\log x)^5-5(\log x)^4+20(\log x)^3-60(\log x)^2+120\right. \\
& (\log x)-120]+C \\
& \therefore \quad A+B+C+D+E+F \\
& =1-5+20-60+120-120=-44 \\
&
\end{aligned}$