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Question Number 26738 by goswamisubhabrata007@gmail.com last updated on 28/Dec/17

Commented by prakash jain last updated on 28/Dec/17

$$\int e^x(\log x+\frac{1}{x})dx$$$$\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx$$$$=\int e^{x}f\left(x\right)dx+\int e^x{f}^{\prime }\left(x\right)dx$$$$=f\left(x\right)e^x-\int e^x{f}^{\prime }\left(x\right)dx+\int e^x{f}^{\prime }\left(x\right)dx$$$$=f\left(x\right)e^x+C$$$$hence$$$$\int e^x(\log x+\frac{1}{x})dx=e^x\log x+C$$

Answered by Penguin last updated on 28/Dec/17

$$\int \frac{e^x}{x}(x\ln \left(x\right)+1)dx=\int e^x\ln \left(x\right)dx+\int \frac{e^x}{x}dx$$$$$$$$\mathrm{solve}\int e^x\ln \left(x\right)dx\mathrm{via}\mathrm{integration}\mathrm{by}\mathrm{parts}$$$$u=e^{x}v=\frac{1}{x}$$$$u^{\prime }=e^x{v}^{\prime }=\ln \left(x\right)$$$$\int e^x\ln \left(x\right)dx=\int {uv}^{\prime }=uv-\int u^{\prime }v$$$$=e^x\ln \left(x\right)-\int \frac{e^x}{x}dx$$$$$$$$\therefore \int e^x\ln \left(x\right)dx+\int \frac{e^x}{x}dx=e^x\ln \left(x\right)+\int \frac{e^x}{x}dx-\int \frac{e^x}{x}dx$$$$\therefore \int \frac{e^x}{x}(x\ln \left(x\right)+1)dx=e^x\ln \left(x\right)+c$$

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