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प्रश्न
Evaluate the following integral:
`int ("d"x)/(2 - 3x - 2x^2)`
उत्तर
`int ("d"x)/(2 - 3x - 2x^2)`
Consider `2 - 3x - 2x^2 = 2[1 - 3/2 x - x^2]`
= `2[1 - (x^2 + 3/2)]`
= `2[1 - ((x + 3/4)^2 - 9/16)]`
= `2[1 + 9/16 - (x + 3/4)^2]`
= `2[25/16 - (x + 3/4)^2]`
= `2[(5/4)^2 - (x + 3/4)^2]`
So integral becomes,
`1/2 int ("d"x)/((5/4)^2 - (x + 3/4)^2) = 1/2 1/(2(5/4)) log |(5/4 + x + 3/4)/(5/4 - x - 3/4)|`
= `1/5 log|(2 + x)/(1 - 2x)| + "c"`
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