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प्रश्न
`int sqrt(4^x(4^x + 4)) "d"x`
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उत्तर
Let I = `int sqrt(4^x(4^x + 4)) "d"x`
= `int sqrt((2^x)^2 [(2^x)^2 + 4]) "d"x`
= `int sqrt((2^x)^2 + 2^2)*2^x "d"x`
Put 2x = t
∴ 2x log2 dx = dt
∴ 2x dx = `1/(log 2) "dt"`
∴ I = `1/(log 2) int sqrt("t"^2 + 2^2) "dt"`
= `1/(log 2)["t"/2 sqrt("t"^2 + 2^2) + 2^2/2log |"t" + sqrt("t"^2 + 2^2)|] + "c"`
∴ I = `1/(log 2) [2^x/2 sqrt(4x + 4) + 2log |2^x + sqrt(4^x + 4)|] + "c"`
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