int log(x + 1) , dx = | vedclass

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(A) To evaluate the integral $I = \int \log(x + 1) \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \log

Vedclass · Accepted Answer

$(x + 1)\log(x + 1) - x + c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \log(x + 1) \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \log(x + 1)$ and $dv = dx$. Then $du = \frac{1}{x + 1} \, dx$ and $v = x$. Applying the formula: $I = x \log(x + 1) - \int \frac{x}{x + 1} \, dx + c$ $I = x \log(x + 1) - \int \frac{x + 1 - 1}{x + 1} \, dx + c$ $I = x \log(x + 1) - \int (1 - \frac{1}{x + 1}) \, dx + c$ $I = x \log(x + 1) - (x - \log(x + 1)) + c$ $I = x \log(x + 1) - x + \log(x + 1) + c$ $I = (x + 1) \log(x + 1) - x + c$.

$\int \log(x + 1) \, dx = $

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