Q112 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
$\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C$
$\Rightarrow 2 K=$
- A$(m+1)(m+2)$
- B$(2-m)(1-m)$
- C$(m+1)(2-m)$
- Correct answerD$(m+2)(1-m)$
Solution
Given that,
$\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}$ ...(i)
Substituting, $\ln x=u$
Differentiating w.r.t ' $x$ ',
$\frac{1}{x} d x=d u$
Putting in Eq. (i)
$\int \frac{d u}{u u^2 u^3 \ldots u^m} \text { or } \int \frac{d u}{u^{1+2+3+\ldots+m}}$
or $\quad \int \frac{d u}{u^{\frac{m(m+1)}{2}}} \quad\left\{1+2+3++n=\frac{n(n+1)}{2}\right\}$
or $\quad \int u^{\frac{-m(m+1)}{2}} d u=\frac{u^{\frac{-m(m+1)}{2}+1}}{\frac{-m(m+1)}{2}+1}+C$
$\left[\because \int x^n d x=\frac{x^{n+1}}{n+1}\right]$
$=\frac{\frac{u(m+2)(1-m)}{2}}{\frac{(m+2)(1-m)}{2}}+C$
Now, putting the value of $u$.
$=\frac{(\ln (x)) \frac{(m+2)(1-m)}{2}}{\frac{(m-2)(1-m)}{2}}+C$
Comparing with given form,
$K=\frac{(m+2)(1-m)}{2} \text { and } 2 K=(m+2)(1-m)$
Hence, the correct answer is option: 'D'.