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प्रश्न
Evaluate the following.
`int "x" sqrt(1 + "x"^2)` dx
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उत्तर
Let I = `int "x" sqrt(1 + "x"^2)` dx
Put 1 + x2 = t
∴ 2x . dx = dt
∴ x . dx = `1/2` dt
∴ I = `1/2 int sqrt"t" * "dt"`
`= 1/2 int "t"^(1/2) *` dt
`= 1/2 * "t"^(3/2)/(3/2)` + c
`= 1/3 "t"^(3/2)` + c
∴ I = `1/3 (1 + "x"^2)^(3/2)` + c
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