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प्रश्न
Integrate the following with respect to x.
`1/(x^2 + 3x + 2)`
उत्तर
`int 1/(x^2 + 3x + 2) "d"x`
Consider `x^2 + 3x + 2 = (x + 3/2)^2 - 9/4 + 2`
= `(x 3/2)^2- 1/4`
=`(x + 3/2)^2 - (1/2)^2`
So the integral becomes
`int ("d"x)/((x + 3/2)^2 - (1/2)^2) = 1/(2(1/2)) log |(x + 3/2 - 1/2)/(x + 3/2 + 1/2)| + "c"`
= `log|(x + 1)/(x +2)| + "c"`
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