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प्रश्न
Integrate the following with respect to x.
`1/sqrt(x^2 + 6x + 13)`
उत्तर
`int 1/sqrt(x^2 + 6x + 13) "d"x`
= `int 1/sqrt((x + 3)^2 + (2)^2) "d"x`
= `log|(x + 3) + sqrt((x + 3)^2 + (2)^2)| + "c"`
= `log|(x + 3) + sqrt(x^2 + 16x + 13)| + "c"`
By Completing the squares
x2 + 6x + 3 = x2 + 6x + (3)2 – (3)2 + 13
= (x + 3)2 – 9 + 13
= (x + 3)2 + 4
= (x + 3)2 + (2)2
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