Q

# Solve it, - Integral Calculus - JEE Main

For x>0,let

• Option 1)

• Option 2)

• Option 3)

• Option 4)

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As learnt in concept

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

 Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable ,$\left ( t\, or\ \theta \right )$

$f(x)=\int_{1}^{x}\frac{logt}{1+t}dx$

$f(\frac{1}{x})=\int_{1}^{\frac{1}{x}}\frac{logt}{1+t}dt$

Put t=$\frac{1}{z}$ (substitution)

=$\int_{1}^{z}\left ( \frac{-log \ z}{1+\frac{1}{z}} \right )\times \frac{-1}{z^{2}}dz$

=$\int_{1}^{z}\left ( \frac{+log \ z}{1+z} \right )\times \frac{1}{z}dz$

$f(x)+f(\frac{1}{x})=\int_{1}^{x}\frac{logt}{1+t}\left ( 1+\frac{1}{t} \right )dt$

$=\int_{1}^{x}\frac{logt}{t} dt$

$=\frac{(logx)^{2}}{2} -0$

Option 1)

This is incorrect option

Option 2)

This is correct option

Option 3)

This is incorrect option

Option 4)

This is incorrect option

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