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Question Number 99123 by Mr.D.N. last updated on 18/Jun/20

Commented by Rasheed.Sindhi last updated on 18/Jun/20

$$Itseemsthat{you}^{\prime }reexperimenting$$$$aboutcolourschemes.Ifyou$$$$allowmetoexpressmypersonal$$$$openionthenIsaythatredtext$$$$onblackbackgroundisnotso$$$$goodformyeyes!$$

Answered by som(math1967) last updated on 18/Jun/20

$$\int \frac{\mathrm{logx}+1}{{(1+\mathrm{logx})}^2}\mathrm{dx}-\int \frac{\mathrm{dx}}{{(1+\mathrm{logx})}^2}$$$$\int \frac{\mathrm{dx}}{(1+\mathrm{logx})}-\int \frac{\mathrm{dx}}{{(1+\mathrm{logx})}^2}$$$$\frac{1}{1+\mathrm{logx}}\int \mathrm{dx}-\int \{\frac{\mathrm{d}}{\mathrm{dx}}\cdot \frac{1}{1+\mathrm{logx}}\int \mathrm{dx}\}\mathrm{dx}-\int \frac{\mathrm{dx}}{{(1+\mathrm{logx})}^2}$$$$\frac{\mathrm{x}}{1+\mathrm{logx}}+\int \frac{1}{{(1+\mathrm{logx})}^2}\cdot \frac{1}{\mathrm{x}}\cdot \mathrm{xdx}$$$$-\int \frac{\mathrm{dx}}{{(1+\mathrm{logx})}^2}$$$$\frac{\mathrm{x}}{1+\mathrm{logx}}+\mathrm{C}\mathrm{ans}$$

Commented by Mr.D.N. last updated on 18/Jun/20

thanks mr.som you are greatest answering.

Answered by Dwaipayan Shikari last updated on 20/Jun/20

$$\int \frac{logx}{{(1+logx)}^2}dx=\int \frac{dx}{(1+logx)}-\int \frac{dx}{{(1+logx)}^2}=\int \frac{xdx}{x(1+logx)}-\int \frac{xdx}{x{(1+logx)}^2}$$$$suppose\left[\right](1+logx)=t$$$$so,\frac{1}{x}=\frac{dt}{dx}$$$$And\left[\right]x=e^{t-1}$$$$This\left[\right]is\left[\right]another\left[\right]form$$$$\int \frac{xdx}{x(1+logx)}-\int \frac{xdx}{x{(1+logx)}^2}=\int e^{t-1}(\frac{1}{t}-\frac{1}{t^2})dt$$$$=\frac{1}{e}\int e^t(\frac{1}{t}-\frac{1}{t^2})dt$$$$=\frac{1}{e}{e}^t\left(\frac{1}{t}\right)+Constant$$$$=\frac{x}{(1+logx)}+Constant$$$$$$$$$$

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