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प्रश्न
`int_0^(log5) (e^x sqrt(e^x - 1))/(e^x + 3) * dx` = ______.
विकल्प
3 + 2π
2 + π
4 – π
4 + π
उत्तर
`int_0^(log5) (e^x sqrt(e^x - 1))/(e^x + 3) * dx` = 4 – π.
Explanation:
Putting ex – 1 = t2 in the given integral, we have
also `d/dx (e^x - 1) = d/dx t^2`
⇒ `e^x = 2t dt/dx`
`int_0^log 5 (e^x sqrt(e^x - 1))/(e^x + 3) dx`
= `2 int_0^2 t^2/(t^2 + 4) dt`
= `2 int_0^2 (t^2 + 4 - 4)/(t^2 + 4) dt`
= `2(int_0^2 1 dt - 4 int_0^2 dt/(t^2 + 4)`
= `2 [(t - 2 tan^(-1) (t/2))_0^2]`
= `2 [(2 - 2 xx pi/4)]`
= 4 – π
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