Mathematics

The value of $$\displaystyle\int { \cfrac { 1 }{ x+x\log { x }  }  } dx$$


ANSWER

$$\log { (1+\log { x } ) } $$


SOLUTION
$$I=\displaystyle\int{\dfrac{dx}{x+x\log{x}}}$$

$$=\displaystyle\int{\dfrac{dx}{x\left(1+\log{x}\right)}}$$

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

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

$$=\log{\left(t\right)}+c$$     .......where $$c$$ is constant of integration

$$=\log{\left(1+\log{x}\right)}+c$$    .........where $$t=1+\log{x}$$
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Single Correct Medium Published on 17th 09, 2020
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