\int \left \{ \frac{(logx-1)}{1+(logx)^{2}} \right \}^{2}dx    is equal to

  • Option 1)

    \frac{x}{x^{2}+1}+C\;

  • Option 2)

    \; \; \frac{logx}{(logx)^{2}+1}+C\;

  • Option 3)

    \; \;\frac{x}{(logx)^{2}+1}+C\;

  • Option 4)

    \; \; \frac{xe^{x}}{1+x^{2}}+C\;

 

Answers (1)
S Sabhrant Ambastha

As we learnt in

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 )

 

 

 and

 

Indefinite integration -

It is inverse process of differentation.

\frac{d}{dx}\left \{ F(x) \right \}= f(x)

\therefore \int f(x)dx= F\left ( x \right )+C

 

- wherein

Where

\frac{d}{dx}F\left ( x \right ) is differential of F(x) w.r.t  x

 

 

I= \int\left [ \frac{logx-1}{1+(logx)^2} \right ]^2dx= \int\frac{(logx-1)^2}{[1+(logx)^2]^2}dx

   =\int\frac{1+(logx)^2-2logx}{[1+(logx)^2]^2}dx

Let  \frac{x}{1+(logx)^2}=t

Differentiating both sides, we get

\frac{[1+(logx)^2]-x[2(logx)\frac{1}{x}]}{[1+(logx)^2]^2}dx=dt

\Rightarrow \frac{1+(logx)^2-2xlogx}{[1+(logx)^2]^2}dx=dt

Thus  I=\int dt= t+c

I=\frac{x}{1+(logx)^2}+c 

 


Option 1)

\frac{x}{x^{2}+1}+C\;

This is incorrect option.

Option 2)

\; \; \frac{logx}{(logx)^{2}+1}+C\;

This is incorrect option.

Option 3)

\; \;\frac{x}{(logx)^{2}+1}+C\;

This is correct option.

Option 4)

\; \; \frac{xe^{x}}{1+x^{2}}+C\;

This is incorrect option.

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