Question Number 226455 by Spillover last updated on 29/Nov/25
Answered by Ghisom_ last updated on 29/Nov/25
![((2−x^2 )/((1−x)(√(1−x^2 ))))=((1+1−x^2 )/((1−x)(√((1−x)(1+x)))))= =(1/((1−x)^(3/2) (1+x)^(1/2) ))+(((1−x)(1+x))/((1−x)^(3/2) (1+x)^(1/2) ))= =(1/((1−x)^(3/2) (1+x)^(1/2) ))+(((1+x)^(1/2) )/((1−x)^(1/2) ))= =(d/dx)[(((1+x)^(1/2) )/((1−x)^(1/2) ))]+(((1+x)^(1/2) )/((1−x)^(1/2) )) we know (d/dx)[f(x)e^x ]=((d/dx)[f(x)]+f(x))e^x ⇒ ∫(((2−x^2 ))/((1−x)(√(1−x^2 ))))e^x dx=(((1+x)^(1/2) )/((1−x)^(1/2) ))e^x +C](https://www.tinkutara.com/question/Q226460.png)
$$\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{\mathrm{1}+\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}}= \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }+\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }+\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} }= \\ $$$$=\frac{{d}}{{dx}}\left[\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} }\right]+\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{know} \\ $$$$\frac{{d}}{{dx}}\left[{f}\left({x}\right)\mathrm{e}^{{x}} \right]=\left(\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]+{f}\left({x}\right)\right)\mathrm{e}^{{x}} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\int\frac{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mathrm{e}^{{x}} {dx}=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} }\mathrm{e}^{{x}} +{C} \\ $$
Commented by Spillover last updated on 30/Nov/25
$${appriciate} \\ $$