Mathematics

Evaluate : $$\displaystyle \int{\dfrac{1}{(x+1)^{2}}}dx$$


SOLUTION
Put $$x+1=t\\dx=dt\\\displaystyle \int \dfrac1{t^2}dt\\\displaystyle \int t^{-2} dt\\\dfrac {t^{-1}}{-1}=-\dfrac 1t +c=\dfrac{-1}{1+x}+c$$
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Subjective Medium Published on 17th 09, 2020
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