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प्रश्न
Integrate the following w.r.t.x : log (log x)+(log x)–2
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उत्तर
Let I = `int [log (logx) + (logx)^-2]*dx`
= `int [log(logx) + 1/(log x)^2]*dx`
Put log x = t
∴ x = et
∴ x = et·dt
∴ I = `int (log t + 1/t^2)e^t*dt`
= `int e^t (log t + 1/t - 1/t + 1/t^2)*dt`
= `int [e^t (log t 1/t) + e^t (-1/t + 1/t^2)]*dt`
= `inte^t (log t + 1/t)*dt - int e^t (1/t - 1/t^2)*dt`
= I1 – I2
In I1, Put f(t) = log t. Then f'(t) = `(1/t)`
∴ I1 = `int e^t [f(t) + f'(t)]*dt`
= `e^t f(t)`
= `e^t log t`
In I2, Put g(t) = `(1/t)`. Then g'(t) = `-(1/t^2)`
∴ I2 = `int e^t ["g"(t) + "g"'(t)]*dt`
= `e^t "g" (t)`
= `e^t*(1/t)`
∴ I = `e^t log t - e^t/t + c`
= `xlog (logx) - x/logx + c`.
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