Question

# Find the value of $\int {x\log (1 - {x^2})} dx$.

Hint: The problem can be solved with the Substitution method. We have to substitute $(1 - {x^2}) = t$. Further, the integration of $\log x$ is $x\log x - x + c$ , where c is the constant of integration.

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

Note: Additional Information, $\int {\log xdx}$ can be calculated by the Integration using the parts.
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.