1.

Evaluate the following definite integrals int_0^1 x/(x^2+1) dx

Answer»

Solution :`x^2+1 = t` Then
dt = 2x dx `gt x dx = dt/2`
ALSO x = 0 `gt t = 1` and
x = 1 `gt t = 2`
THEREFORE `int_0^1 x/(x^2+1) dx = int_1^2 1/t dt/2`
=`1/2[LOG |t|]_1^2`
=`1/2[log|2| -log|1|]`
`=1/2(log 2-0) = 1/2 log2`


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