Mathematics

If $$I=\int e^x\frac{(1-x)^2}{(1+x^2)^2}dx $$ , then I =


ANSWER

$$ e^x\frac{1}{(1+x^2)}+C$$


SOLUTION

 $$I=\int e^x\frac{(1-x)^2}{(1+x^2)^2}dx $$
$$I=\int e^x\frac{(1+x^2-2x)}{(1+x^2)^2}dx $$
$$I=\int e^x\frac{1+x^2}{(1+x^2)^2}dx $$ $$-$$ $$\int e^x\frac{2x}{(1+x^2)^2}dx $$ 
$$I=\int e^x\frac{1}{(1+x^2)}dx $$ $$-$$ $$\int e^x\frac{2x}{(1+x^2)^2}dx $$ 
By integration by parts : $$\int [f(x) g(x)]dx=f(x)\cdot \int g(x)dx-\int[f'(x) \int g(x) dx]dx$$
By integration by  parts of first integral
$$I=$$ $$\frac { 1 }{ 1+{ x }^{ 2 } } \int { e } ^{ x }-\int { \{ (\frac { d(\frac { 1 }{ 1+{ x }^{ 2 } } ) }{ dx }  } )\int { { e }^{ x } } dx\} dx$$   $$-$$ $$\int e^x\frac{2x}{(1+x^2)^2}dx $$
$$I= e^x\frac{1}{(1+x^2)} $$ $$+$$ $$\int e^x\frac{2x}{(1+x^2)^2}dx $$ $$-$$ $$\int e^x\frac{2x}{(1+x^2)^2}dx $$
$$I= e^x\frac{1}{(1+x^2)} $$ $$+$$ $$C$$
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Single Correct Medium Published on 17th 09, 2020
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