The value of the integral int_{-1}^2 log _ele | vedclass

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Question

(B) Let $I = \int_{-1}^2 1 \cdot \log _e(x+\sqrt{x^2+1}) dx$.Using integration by parts,$\int u dv = uv - \int v du$,where $u = \log _e(x+\sqrt{x^2+1}

Vedclass · Accepted Answer

$\sqrt{2}-\sqrt{5}+\log _{e}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$

Vedclass · Answer

(B) Let $I = \int_{-1}^2 1 \cdot \log _e(x+\sqrt{x^2+1}) dx$.Using integration by parts,$\int u dv = uv - \int v du$,where $u = \log _e(x+\sqrt{x^2+1})$ and $dv = dx$.Then $du = \frac{1}{x+\sqrt{x^2+1}} \cdot (1 + \frac{x}{\sqrt{x^2+1}}) dx = \frac{1}{\sqrt{x^2+1}} dx$ and $v = x$.$I = [x \log _e(x+\sqrt{x^2+1})]_{-1}^2 - \int_{-1}^2 \frac{x}{\sqrt{x^2+1}} dx$.$I = [x \log _e(x+\sqrt{x^2+1}) - \sqrt{x^2+1}]_{-1}^2$.$I = (2 \log _e(2+\sqrt{5}) - \sqrt{5}) - (-1 \log _e(-1+\sqrt{2}) - \sqrt{2})$.$I = 2 \log _e(2+\sqrt{5}) + \log _e(\sqrt{2}-1) - \sqrt{5} + \sqrt{2}$.Since $\log _e(\sqrt{2}-1) = \log _e(\frac{1}{\sqrt{2}+1}) = -\log _e(\sqrt{2}+1)$,we have:$I = \log _e(2+\sqrt{5})^2 - \log _e(\sqrt{2}+1) - \sqrt{5} + \sqrt{2}$.$I = \sqrt{2} - \sqrt{5} + \log _e\left(\frac{9+4\sqrt{5}}{\sqrt{2}+1}\right)$.

The value of the integral $\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ is:

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