Mathematics

# Solve $\int {\dfrac{1}{{x\sqrt {{x^2} + x + 1} }}} dx$

##### SOLUTION

Given

$\int\dfrac{1}{x\sqrt{x^2+x+1}}dx$

$\int\dfrac{1}{x^2\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}$

$Put \dfrac{1}{x}=t, than\dfrac{1}{x^2}dx=-dt$

Now,

$\int\dfrac{-1}{\sqrt{t^2+t+1}}dt$

$\int\dfrac{-1}{\sqrt{t^2+t+\dfrac{1}{4}+\dfrac{3}{4}}}$

$\int\dfrac{-1}{\sqrt{\left(t+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}}$

$\Rightarrow-\log{t+\sqrt{t^2+\dfrac{3}{4}}}+constant$

$\Rightarrow-\log\left(\dfrac{1}{x}+\sqrt{\dfrac{1}{x^2}+\dfrac{3}{4}}\right)+constant$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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