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CUET
-- Mathematics - Section A
Definite Integration
If \(\int_{0}^{1}\frac{e^t}{1+t}dt=a\), then \(\int_{0}^{1}\frac{e^t}{(1+t)^2}dt=\)
\(a-1+\frac{e}{2}\)
\(a-1-\frac{e}{2}\)
\(a+1-\frac{e}{2}\)
\(a+1+\frac{e}{2}\)