int_0^1 log (x+1) , dx = | vedclass

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(A) To evaluate the integral $I = \int_0^1 \log(x+1) \, dx$,we use the method of integration by parts.Let $u = \log(x+1)$ and $dv = dx$.Then $du = \fr

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$2 \log 2 - 1$

Vedclass · Answer

(A) To evaluate the integral $I = \int_0^1 \log(x+1) \, dx$,we use the method of integration by parts.Let $u = \log(x+1)$ and $dv = dx$.Then $du = \frac{1}{x+1} \, dx$ and $v = x$.Using the formula $\int u \, dv = uv - \int v \, du$,we get:$I = [x \log(x+1)]_0^1 - \int_0^1 \frac{x}{x+1} \, dx$$I = [1 \cdot \log(2) - 0 \cdot \log(1)] - \int_0^1 \frac{x+1-1}{x+1} \, dx$$I = \log 2 - \int_0^1 (1 - \frac{1}{x+1}) \, dx$$I = \log 2 - [x - \log(x+1)]_0^1$$I = \log 2 - [(1 - \log 2) - (0 - \log 1)]$$I = \log 2 - 1 + \log 2$$I = 2 \log 2 - 1$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$\int_0^1 \log (x+1) \, dx =$

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