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CUET
-- Mathematics - Section A
Indefinite Integration
$∫e^x\left(\frac{1-x}{1+x^2}\right)^2dx=$
$\frac{e^x}{1+x^2}+C$
$-\frac{e^x}{1+x^2}+C$
$\frac{e^x}{(1+x^2)^2}+C$
$-\frac{e^x}{(1+x^2)^2}+C$