ANSWER on Question #72783 Math. Complex Analysis
Simplify
∫ 1 2 1 + 1 t 2 + ( ln t ) 2 + 2 ln t + 1 d t \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (\ln t)^{2} + 2 \ln t + 1} \, dt ∫ 1 2 1 + t 2 1 + ( ln t ) 2 + 2 ln t + 1 d t
SOLUTION
∫ 1 2 1 + 1 t 2 + ( ln t ) 2 + 2 ln t + 1 d t = ∫ 1 2 1 + 1 t 2 + ( ( ln t ) 2 + 2 ln t + 1 ) ⏟ ( ln t + 1 ) 2 d t = = ∫ 1 2 1 + 1 t 2 + ( 1 + ln t ) 2 d t = ∫ 1 2 t 2 + 1 + t 2 ⋅ ( 1 + ln t ) 2 t 2 d t = = ∫ 1 2 t 2 + 1 + t 2 ⋅ ( 1 + ln t ) 2 t d t ≡ ∫ 1 2 t 2 + 1 + t 2 ⋅ ( 1 + ln t ) 2 d t t = = [ ln t = k → d t t = d k t = e k t = 1 → k = ln 1 = 0 t = 2 → k = ln 2 ] = ∫ 0 ln 2 ( e k ) 2 + 1 + ( e k ) 2 ⋅ ( 1 + k ) 2 d k = = ∫ 0 ln 2 e 2 k ⋅ ( ( 1 + k ) 2 + 1 ) + 1 d k \begin{aligned}
\int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (\ln t)^{2} + 2 \ln t + 1} \, dt &= \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + \underbrace{((\ln t)^{2} + 2 \ln t + 1)}_{(\ln t + 1)^{2}}} \, dt = \\
&= \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (1 + \ln t)^{2}} \, dt = \int_{1}^{2} \sqrt{\frac{t^{2} + 1 + t^{2} \cdot (1 + \ln t)^{2}}{t^{2}}} \, dt = \\
&= \int_{1}^{2} \frac{\sqrt{t^{2} + 1 + t^{2} \cdot (1 + \ln t)^{2}}}{t} \, dt \equiv \int_{1}^{2} \sqrt{t^{2} + 1 + t^{2} \cdot (1 + \ln t)^{2}} \, \frac{dt}{t} = \\
&= \left[ \begin{array}{c} \ln t = k \to \frac{dt}{t} = dk \\ t = e^{k} \\ t = 1 \to k = \ln 1 = 0 \\ t = 2 \to k = \ln 2 \end{array} \right] = \int_{0}^{\ln 2} \sqrt{(e^{k})^{2} + 1 + (e^{k})^{2} \cdot (1 + k)^{2}} \, dk = \\
&= \int_{0}^{\ln 2} \sqrt{e^{2k} \cdot ((1 + k)^{2} + 1) + 1} \, dk
\end{aligned} ∫ 1 2 1 + t 2 1 + ( ln t ) 2 + 2 ln t + 1 d t = ∫ 1 2 1 + t 2 1 + ( l n t + 1 ) 2 (( ln t ) 2 + 2 ln t + 1 ) d t = = ∫ 1 2 1 + t 2 1 + ( 1 + ln t ) 2 d t = ∫ 1 2 t 2 t 2 + 1 + t 2 ⋅ ( 1 + ln t ) 2 d t = = ∫ 1 2 t t 2 + 1 + t 2 ⋅ ( 1 + ln t ) 2 d t ≡ ∫ 1 2 t 2 + 1 + t 2 ⋅ ( 1 + ln t ) 2 t d t = = ⎣ ⎡ ln t = k → t d t = d k t = e k t = 1 → k = ln 1 = 0 t = 2 → k = ln 2 ⎦ ⎤ = ∫ 0 l n 2 ( e k ) 2 + 1 + ( e k ) 2 ⋅ ( 1 + k ) 2 d k = = ∫ 0 l n 2 e 2 k ⋅ (( 1 + k ) 2 + 1 ) + 1 d k
ANSWER
∫ 1 2 1 + 1 t 2 + ( ln t ) 2 + 2 ln t + 1 d t = ∫ 0 ln 2 e 2 k ⋅ ( ( 1 + k ) 2 + 1 ) + 1 d k \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (\ln t)^{2} + 2 \ln t + 1} \, dt = \int_{0}^{\ln 2} \sqrt{e^{2k} \cdot ((1 + k)^{2} + 1) + 1} \, dk ∫ 1 2 1 + t 2 1 + ( ln t ) 2 + 2 ln t + 1 d t = ∫ 0 l n 2 e 2 k ⋅ (( 1 + k ) 2 + 1 ) + 1 d k
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#340153 on Dec 2023