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
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