Question

integration of 1 by 1+X dx
​

Answer 1

$\textbf{Given:}$

$\int\,\dfrac{1}{1+x}\,dx$

$\textbf{Solution:}$

$\text{We apply change of variable method to solve this integral}$

$\text{Consider,}$

$\int\,\dfrac{1}{1+x}\,dx$

$\text{Take t=1+x }$

$\dfrac{dt}{dx}=1$

$\implies\,dt=dx$

$\text{Now,}$

$\int\,\dfrac{1}{1+x}\,dx$

$=\int\,\dfrac{1}{t}\,dt$

$\text{Using the formula,}$

$\boxed{\bf\int\,\dfrac{1}{1x}\,dx=\log|x|+c}$

$\text{we get}$

$=\log|t|+c$

$=\log|1+x|+c$

$\therefore\bf\int\,\dfrac{1}{1+x}\,dx=\log|1+x|+c$

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