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प्रश्न
Find the integrals of the following:
`1/sqrt(xx^2 + 4x + 2)`
उत्तर
`int ("d"x)/sqrt(x^2 + 4x + 2) = int ("d"x)/sqrt((x + 2)^2 - 2^2 + 2)`
= `int ("d"x)/sqrt((x + 2)^2 - 4 + 2)`
= `int ("d"x)/sqrt((x + 2)^2 - 2)`
= `int ("d"x)/sqrt((x + 2)^2 - (sqrt(2))^2`
Put x + 2 = t
dx = dt
= `int ("d"x)/sqrt("t"^2 - (sqrt(2))^2`
`int ("d"x)/sqrt(x^2 - "a"^2) = log|x + sqrt(x^2 - "a"^2)| + "c"`
= `log |"t" + sqrt("t"^2 - (sqrt(2))^2)| + "c"`
= `log |(x + 2) + sqrt((x + 2)^2 - 2)| + "c"`
= `log |(x + 2) + sqrt(x^2 + 4x + 4 - 2)| + "c"`
`int ("d"x)/sqrt(x^2 + 4x + 2) = log|(x + 2) + sqrt(x^2 + 4x + 2)| + "c"`
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