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

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(B) Let $I = \int (1+x) \log x \, dx$. Using integration by parts,let $u = \log x$ and $dv = (1+x) \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = x +

Vedclass · Accepted Answer

$\left(x+\frac{x^{2}}{2}\right) \log x-\left(x+\frac{x^{2}}{4}\right)+C$

Vedclass · Answer

(B) Let $I = \int (1+x) \log x \, dx$. Using integration by parts,let $u = \log x$ and $dv = (1+x) \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = x + \frac{x^2}{2}$. Using the formula $\int u \, dv = uv - \int v \, du$: $I = \log x \left(x + \frac{x^2}{2}\right) - \int \left(x + \frac{x^2}{2}\right) \cdot \frac{1}{x} \, dx$. $I = \left(x + \frac{x^2}{2}\right) \log x - \int \left(1 + \frac{x}{2}\right) \, dx$. $I = \left(x + \frac{x^2}{2}\right) \log x - \left(x + \frac{x^2}{4}\right) + C$.

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

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