Mathematics

Evaluate $$\displaystyle\int_{1}^{2}\dfrac{1}{x(1+\log x)^{2}}dx$$


SOLUTION
$$\displaystyle\int_{1}^{2}{\dfrac{1}{x{\left(1+\log{x}\right)}^{2}}dx}$$

Let $$t=1+\log{x}\Rightarrow\,dt=\dfrac{1}{x}dx$$

$$=\displaystyle\int{\dfrac{dt}{{t}^{2}}}$$

$$=\displaystyle\int{{t}^{-2}dt}$$

$$=\left[\dfrac{{t}^{-2+1}}{-2+1}\right]_{1}^{2}$$

$$=\left[-\dfrac{1}{t}\right]$$

$$=\left[-\dfrac{1}{1+\log{x}}\right]_{1}^{2}$$ where $$t=1+\log{x}$$

$$=-\left[\dfrac{1}{1+\log{x}}\right]_{1}^{2}$$

$$=-\left[\dfrac{1}{1+\log{2}}-\dfrac{1}{1+\log{1}}\right]$$

$$=-\left[\dfrac{1}{1+\log{2}}-\dfrac{1}{1+0}\right]$$

$$=-\left[\dfrac{1}{1+\log{2}}-1\right]$$

$$=-\left[\dfrac{1-1-\log{2}}{1+\log{2}}\right]$$

$$=-\left[\dfrac{-\log{2}}{1+\log{2}}\right]$$

$$=\dfrac{\log{2}}{1+\log{2}}$$

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Subjective Medium Published on 17th 09, 2020
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