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