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प्रश्न
Evaluate: `int_1^2 x/(1 + x^2) "d"x`
उत्तर
`int_1^2 x/(1 + x^2) "d"x = 1/2 int_1^2 (2x)/(1 + x^2) "d"x`
= `1/2[log|1 + x^2|]_1^2` ........`[∵ int ("f'"(x))/("f"(x)) "d"x = log|"f"(x)| + c"]`
= `1/2(log 5 - log 2)`
=`1/2 log(5/2)`
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