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Substitute t at the place of $(1 - {x^2})$ in $\int {x\log (1 - {x^2})} dx$

If, $(1 - {x^2}) = t$

Then, differentiating both sides with respect to $t$, we get

$

\Rightarrow \dfrac{{d(1 - {x^2})}}{{dt}} = dt \\

\Rightarrow - 2xdx = dt \\

\Rightarrow xdx = - \dfrac{{dt}}{2} \\

$

Putting above value in $\int {x\log (1 - {x^2})} dx$, we get

$

\Rightarrow \int { - \log t\dfrac{{dt}}{2}} \\

\Rightarrow - \dfrac{1}{2}\int {\log tdt} \\

$

Further we know $\int {\log t = t\log t - t + c} $ where c is the constant of integration.

$ \Rightarrow - \dfrac{1}{2}\int {\log tdt} = - \dfrac{1}{2}(t\log t - t + c)$

Putting the value $t = (1 - {x^2})$ in above equation, we get

$

\Rightarrow - \dfrac{1}{2}(t\log t - t + c) \\

\Rightarrow - \dfrac{1}{2}((1 - {x^2})\log (1 - {x^2}) - (1 - {x^2}) + c) \\

$

As, We know that $\int {f(x)g(x)dx = f(x)\int {g(x)} } dx - \int {\left( {f'(x} \right)} \int {g\left( x \right)} dx)dx$

Here, $f(x) = \log x$ and $g(x) = 1$

$

\Rightarrow \int {(\log x)} 1dx = \log x\int {1.dx - \int {(\dfrac{{d(\log x)}}{{dx}}} } \int {1.dx)dx} \\

= (\log x)x - \int {\dfrac{1}{x}} .x.dx \\

= x\log x - \int {1.dx} \\

= x\log x - x + c \\

$

where c is the constant of integration.

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