Mathematics

Integrate:
$$\displaystyle \int { \dfrac { 1 }{ x\sqrt { x+1 }  }  } dx$$


SOLUTION
$$\displaystyle  \int \dfrac{1}{x\sqrt{x+1}}dx$$
Let $$ \sqrt{x+1} = t $$
$$ x+1 = t^{2}$$
$$ x = t^{2}-1$$
$$ \dfrac{dx}{dt} = \dfrac{(dt^{2}-1)}{dt} =2t$$
$$ dx = 2t dt $$
$$ \int \dfrac{1}{(t^{2}-1)t} 2t.dt$$
$$ = \int \dfrac{2dt}{(t^{2}-1)}$$
$$ = \int \dfrac{2dt}{(t+1)(t-1)}$$
$$ = \int (\dfrac{1}{t-1}-\dfrac{1}{t+1})dt$$
$$ = \int \dfrac{dt}{t-1}-\int \dfrac{dt}{t+1}$$
$$ = \left | n \right |\left | t-1 \right |-\left | n \right |\left | t+1 \right |$$
$$ = ln \left | \dfrac{t-1}{t+1} \right |$$
$$ = ln \left | \dfrac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right |$$ Answers
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Subjective Medium Published on 17th 09, 2020
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