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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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