$\int\left\{\frac{(\log x-1)}{\left(1+(\log x)^{2}\right.}\right\}^{2} d x$ is equal to

(D) $\frac{x}{(\log x)^{2}+1}+C$

(A) $\frac{\log x}{(\log x)^{2}+1}+C$

(C) $\frac{x e^{x}}{1+x^{2}}+C$

JEE Main · 2005 · 08 May 2005 · Offline · Indefinite Integrals

\int {\frac{( l o g x - 1 )}{( 1 + ( l o g x ) ^{2}}}^{2} d x

is equal to

Answer: (D)

Expand the square to get

\int \frac{( ln x - 1 ) ^{2}}{( 1 + ( ln x ) ^{2} ) ^{2}} d x = \int [\frac{1}{1 + ( ln x ) ^{2}} - \frac{2 ln x}{( 1 + ( ln x ) ^{2} ) ^{2}}] d x

. With

t = ln x

and

d x = e^{t} d t

, this is

\int e^{t} [\frac{1}{1 + t ^{2}} - \frac{2 t}{( 1 + t ^{2} ) ^{2}}] d t

. The bracket is

g (t) + g^{'} (t)

with

g (t) = \frac{1}{1 + t ^{2}}

, since

g^{'} (t) = - \frac{2 t}{( 1 + t ^{2} ) ^{2}}

, and

\int e^{t} [g (t) + g^{'} (t)] d t = e^{t} g (t) + C

. Hence the integral equals

\frac{e ^{t}}{1 + t ^{2}} + C = \frac{x}{1 + ( ln x ) ^{2}} + C

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option