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Question Number 135646 by metamorfose last updated on 14/Mar/21

$$\int (x+\frac{1}{2})ln(1+\frac{1}{x})-xdx=...?$$

Answered by Ñï= last updated on 15/Mar/21

$$\int [(x+\frac{1}{2})ln(1+\frac{1}{x})-x]dx$$$$=\int [(x+\frac{1}{2})(ln(x+1)-lnx)-x]dx$$$$=\int (x+\frac{1}{2})ln(x+1)dx-\int (x+\frac{1}{2})lnxdx-\int xdx$$$$=(\frac{1}{2}{x}^2+\frac{1}{2}x)ln(x+1)-\int \frac{x+1-\frac{1}{2}}{x+1}dx-(\frac{1}{2}{x}^2+\frac{1}{2}x)lnx+\int \frac{x+\frac{1}{2}}{x}dx-\frac{1}{2}{x}^2$$$$=(\frac{1}{2}{x}^2+\frac{1}{2}x)ln(1+\frac{1}{x})+\frac{1}{2}ln(x^2+x)-\frac{1}{2}{x}^2+C$$

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