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

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.

Posted by

Sabhrant Ambastha

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