Let F( x ) = f( x ) + f( 1 \over x ), where f( x ) = \int\limits_l^x \log t \over 1 + tdt, Then F(e) equals

Definite Integration PYQ — JEE Main 2007 · Shift 0 · Q49 with Solution

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Correct answer: C

Given functions

$f(x)=\int_1^x \frac{\log t}{1+t}\,dt$ and $F(x)=f(x)+f\left(\frac1x\right).$

We need to find $F(e)=f(e)+f\left(\frac1e\right).$
Write both integrals explicitly

$f(e)=\int_1^e \frac{\log t}{1+t}\,dt$ and $f\left(\frac1e\right)=\int_1^{1/e} \frac{\log t}{1+t}\,dt.$

Hence, $F(e)=\int_1^e \frac{\log t}{1+t}\,dt+\int_1^{1/e} \frac{\log t}{1+t}\,dt.$
Transform the second integral

Consider $I=\int_1^{1/e} \frac{\log t}{1+t}\,dt.$

Put $t=\frac1u \quad \Rightarrow \quad dt=-\frac1{u^2}du.$

Also,
$\qquad 1+t=1+\frac1u=\frac{u+1}{u}.$$ Therefore, $$\frac{\log t}{1+t}dt= rac{-\log u}{(u+1)/u}\left(-\frac1{u^2}du\right) =\frac{\log u}{u(u+1)}du.$$ Limits change as: - when $t=1$, $u=1$ - when $t=1/e$, $u=e$ So, $$I=\int_1^e \frac{\log u}{u(u+1)}du.$$$
Now add the two parts

Thus, $F(e)=\int_1^e \frac{\log t}{1+t}dt+\int_1^e \frac{\log t}{t(1+t)}dt.$

Combine them: $F(e)=\int_1^e \log t\left(\frac1{1+t}+\frac1{t(1+t)}\right)dt.$

Simplify inside:
$=\frac{t}{t(1+t)}+\frac1{t(1+t)} =\frac{t+1}{t(1+t)} =\frac1t.$$ Hence, $$F(e)=\int_1^e \frac{\log t}{t}dt.$$$
Evaluate the integral

We know $\int \frac{\log t}{t}dt=\frac{(\log t)^2}{2}.$

Therefore,
$=\frac{(\log e)^2}{2}-\frac{(\log 1)^2}{2} =\frac{1^2}{2}-0 =\frac12.$$$
Check options

$F(e)=\frac12$

So the correct option is C.

View written solutionFree

Correct answer: C

Given functions

$f(x)=\int_1^x \frac{\log t}{1+t}\,dt$ and $F(x)=f(x)+f\left(\frac1x\right).$

We need to find $F(e)=f(e)+f\left(\frac1e\right).$
Write both integrals explicitly

$f(e)=\int_1^e \frac{\log t}{1+t}\,dt$ and $f\left(\frac1e\right)=\int_1^{1/e} \frac{\log t}{1+t}\,dt.$

Hence, $F(e)=\int_1^e \frac{\log t}{1+t}\,dt+\int_1^{1/e} \frac{\log t}{1+t}\,dt.$
Transform the second integral

Consider $I=\int_1^{1/e} \frac{\log t}{1+t}\,dt.$

Put $t=\frac1u \quad \Rightarrow \quad dt=-\frac1{u^2}du.$

Also,
$\qquad 1+t=1+\frac1u=\frac{u+1}{u}.$$ Therefore, $$\frac{\log t}{1+t}dt= rac{-\log u}{(u+1)/u}\left(-\frac1{u^2}du\right) =\frac{\log u}{u(u+1)}du.$$ Limits change as: - when $t=1$, $u=1$ - when $t=1/e$, $u=e$ So, $$I=\int_1^e \frac{\log u}{u(u+1)}du.$$$
Now add the two parts

Thus, $F(e)=\int_1^e \frac{\log t}{1+t}dt+\int_1^e \frac{\log t}{t(1+t)}dt.$

Combine them: $F(e)=\int_1^e \log t\left(\frac1{1+t}+\frac1{t(1+t)}\right)dt.$

Simplify inside:
$=\frac{t}{t(1+t)}+\frac1{t(1+t)} =\frac{t+1}{t(1+t)} =\frac1t.$$ Hence, $$F(e)=\int_1^e \frac{\log t}{t}dt.$$$
Evaluate the integral

We know $\int \frac{\log t}{t}dt=\frac{(\log t)^2}{2}.$

Therefore,
$=\frac{(\log e)^2}{2}-\frac{(\log 1)^2}{2} =\frac{1^2}{2}-0 =\frac12.$$$
Check options

$F(e)=\frac12$

So the correct option is C.