Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

$∫e^x\left(\frac{1-x}{1+x^2}\right)^2dx=$

Options:

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

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

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

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

Correct Answer:

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