Question

Q. simplify ∫_1^2▒√(1+1/t^2 +〖(lnt)〗^2+2lnt+1)dt

Answer 1

ANSWER on Question #72783 Math. Complex Analysis

Simplify

$$\int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (\ln t)^{2} + 2 \ln t + 1} \, dt$$

SOLUTION

$$\begin{aligned} \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (\ln t)^{2} + 2 \ln t + 1} \, dt &= \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + \underbrace{((\ln t)^{2} + 2 \ln t + 1)}_{(\ln t + 1)^{2}}} \, dt = \\ &= \int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (1 + \ln t)^{2}} \, dt = \int_{1}^{2} \sqrt{\frac{t^{2} + 1 + t^{2} \cdot (1 + \ln t)^{2}}{t^{2}}} \, dt = \\ &= \int_{1}^{2} \frac{\sqrt{t^{2} + 1 + t^{2} \cdot (1 + \ln t)^{2}}}{t} \, dt \equiv \int_{1}^{2} \sqrt{t^{2} + 1 + t^{2} \cdot (1 + \ln t)^{2}} \, \frac{dt}{t} = \\ &= \left[ \begin{array}{c} \ln t = k \to \frac{dt}{t} = dk \\ t = e^{k} \\ t = 1 \to k = \ln 1 = 0 \\ t = 2 \to k = \ln 2 \end{array} \right] = \int_{0}^{\ln 2} \sqrt{(e^{k})^{2} + 1 + (e^{k})^{2} \cdot (1 + k)^{2}} \, dk = \\ &= \int_{0}^{\ln 2} \sqrt{e^{2k} \cdot ((1 + k)^{2} + 1) + 1} \, dk \end{aligned}$$

ANSWER

$$\int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + (\ln t)^{2} + 2 \ln t + 1} \, dt = \int_{0}^{\ln 2} \sqrt{e^{2k} \cdot ((1 + k)^{2} + 1) + 1} \, dk$$

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