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Question Number 113278 by mathdave last updated on 12/Sep/20

s o l v e

\int \frac{\sqrt{x^{2} + x + 2 - \sqrt{4 x^{2} + 4 x + 4}}}{x \sqrt{x^{4} + x^{3} + x^{2}}} d x

Answered by 1549442205PVT last updated on 13/Sep/20

x^{2} + x + 2 - \sqrt{{4 x}^{2} + 4 x + 4} =

{(\sqrt{x^{2} + x + 1})}^{2} - 2 \sqrt{x^{2} + x + 1} + 1

= {(\sqrt{x^{2} + x + 1} - 1)}^{2}, x \sqrt{x^{4} + x^{3} + x^{2}} = x^{2} \sqrt{x^{2} + x + 1}

Hence, F = \int \frac{\sqrt{x^{2} + x + 2 - \sqrt{4 x^{2} + 4 x + 4}}}{x \sqrt{x^{4} + x^{3} + x^{2}}} d x

= \int \frac{\sqrt{x^{2} + x + 1} - 1}{x^{2} \sqrt{x^{2} + x + 1}} dx = \int (\frac{1}{x^{2}} - \frac{1}{x^{2} \sqrt{x^{2} + x + 1}}) dx

We find J = \int \frac{dx}{x^{2} \sqrt{x^{2} + x + 1}} = \int \frac{\sqrt{x^{2} + x + 1}}{x^{2} (x^{2} + x + 1)} dx

= \int \sqrt{x^{2} + x + 1} (\frac{- x + 1}{x^{2}} + \frac{x}{x^{2} + x + 1}) dx

= \int (\frac{\sqrt{x^{2} + x + 1}}{x^{2}} - \frac{\sqrt{x^{2} + x + 1}}{x} + \frac{x}{\sqrt{x^{2} + x + 1}}) dx

== \int [\frac{\sqrt{x^{2} + x + 1}}{x^{2}} - \frac{\sqrt{x^{2} + x + 1}}{x} - \frac{1}{2 \sqrt{x^{2} + x + 1}} + (\frac{1}{2} \times \frac{2 x + 1}{\sqrt{x^{2} + x + 1}})] dx

\frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x + 1}} dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} (u = x^{2} + x + 1)

= \sqrt{x^{2} + x + 1}

Since \int \frac{dx}{\sqrt{x^{2} + λ}} = \ln ∣ x + \sqrt{x^{2} + λ} ∣ + C

\frac{1}{2} \int \frac{dx}{\sqrt{x^{2} + x + 1}} = \frac{1}{2} \int \frac{d (x + \frac{1}{2})}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}} = \frac{1}{2} \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣

Therefore, we just need find

= \int (\frac{\sqrt{x^{2} + x + 1}}{x^{2}} - \frac{\sqrt{x^{2} + x + 1}}{x})] dx = A + B

A = \int \frac{\sqrt{x^{2} + x + 1}}{x^{2}} = - \int \sqrt{x^{2} + x + 1} d (\frac{1}{x})

\underset{by part}{=} \frac{- \sqrt{x^{2} + x + 1}}{x} + \int \frac{1}{x} \times \frac{2 x + 1}{2 \sqrt{x^{2} + x + 1}} dx

= \frac{- \sqrt{x^{2} + x + 1}}{x} + \int \frac{dx}{\sqrt{x^{2} + x + 1}} + \int \frac{dx}{2 x \sqrt{x^{2} + x + 1}}

= \frac{- \sqrt{x^{2} + x + 1}}{x} + \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣ + \frac{1}{2} M (1)

B = \int \frac{\sqrt{x^{2} + x + 1}}{x} dx \underset{by parts}{=} \sqrt{x^{2} + x + 1} -

- \int [x . \frac{. (\frac{(2 x + 1) x}{2 \sqrt{x^{2} + x + 1}} - \sqrt{x^{2} + x + 1})}{x^{2}}] dx

= \sqrt{x^{2} + x + 1} - \int \frac{- x - 2}{2 x \sqrt{x^{2} + x + 1}} dx

= \sqrt{x^{2} + x + 1} + \int \frac{dx}{2 \sqrt{x^{2} + x + 1}} + \int \frac{dx}{x \sqrt{x^{2} + x + 1}}

= \sqrt{x^{2} + x + 1} + \frac{1}{2} \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣ + M (2)

M = \int \frac{dx}{x \sqrt{x^{2} + x + 1}} = \int (\frac{2}{2 x \sqrt{x^{2} + x + 1}} \times \frac{2 x + 1 - 2 \sqrt{x^{2} + x + 1}}{2 x + 1 - 2 \sqrt{x^{2} + x + 1}}) dx

= \int (\frac{2 x + 1 - 2 \sqrt{x^{2} + x + 1}}{2 \sqrt{x^{2} + x + 1}} \times \frac{2}{x (2 x + 1) - 2 \sqrt{x^{2} + x + 1}}) dx

= \int (\frac{2 x + 1}{2 \sqrt{x^{2} + x + 1}} - 1) (\frac{1}{\sqrt{x^{2} + x + 1} - x - 1} - \frac{1}{\sqrt{x^{2} + x + 1} - x + 1}) dx

= \int (\frac{\frac{2 x + 1}{2 \sqrt{x^{2} + x + 1}} - 1}{\sqrt{x^{2} + x + 1} - x - 1} - \frac{\frac{2 x + 1}{2 \sqrt{x^{2} + x + 1}} - 1}{\sqrt{x^{2} + x + 1} - x + 1}) dx

= \ln ∣ \sqrt{x^{2} + x + 1} - x - 1 ∣ - \ln ∣ \sqrt{x^{2} + x + 1} - x + 1 ∣

= \ln ∣ \frac{\sqrt{x^{2} + x + 1} - x - 1}{\sqrt{x^{2} + x + 1} - x + 1} ∣ (3)

From (1) (2) (3) we get

A + B = - \frac{\sqrt{x^{2} + x + 1}}{x} + \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣

+ \sqrt{x^{2} + x + 1} + \frac{1}{2} \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣

+ \frac{3}{2} \ln ∣ \frac{\sqrt{x^{2} + x + 1} - x - 1}{\sqrt{x^{2} + x + 1} - x + 1} ∣ (4)

Therefore, J = \sqrt{x^{2} + x + 1} - \frac{1}{2} \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣

+ A + B = 2 \sqrt{x^{2} + x + 1} + \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣ +

+ \frac{3}{2} \ln ∣ \frac{\sqrt{x^{2} + x + 1} - x - 1}{\sqrt{x^{2} + x + 1} - x + 1} ∣ . From that

F = - \frac{1}{x} + J = - \frac{1}{x} + A + B = 2 \sqrt{x^{2} + x + 1} + \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣ +

+ \frac{3}{2} \ln ∣ \frac{\sqrt{x^{2} + x + 1} - x - 1}{\sqrt{x^{2} + x + 1} - x + 1} ∣ + C (C - constant)

Thus, final result is

F = \int \frac{\sqrt{x^{2} + x + 2 - \sqrt{4 x^{2} + 4 x + 4}}}{x \sqrt{x^{4} + x^{3} + x^{2}}} d x

= - \frac{1}{x} + 2 \sqrt{x^{2} + x + 1} + \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x + 1} ∣

+ \frac{3}{2} \ln ∣ \frac{\sqrt{x^{2} + x + 1} - x - 1}{\sqrt{x^{2} + x + 1} - x + 1} ∣ + C

Commented by mathdave last updated on 12/Sep/20

n i c e o n e

Commented by 1549442205PVT last updated on 13/Sep/20

I like smart {Sir}^{'} s solution

Commented by Tawa11 last updated on 06/Sep/21

grest sir

Answered by MJS_new last updated on 12/Sep/20

\int \frac{\sqrt{x^{2} + x + 1 - 2 \sqrt{x^{2} + x + 1} + 1}}{x^{2} \sqrt{x^{2} + x + 1}} d x =

= \int \frac{d x}{x^{2}} - \int \frac{d x}{x^{2} \sqrt{x^{2} + x + 1}}

(1) \int \frac{d x}{x^{2}} = - \frac{1}{x}

(2) - \int \frac{d x}{x^{2} \sqrt{x^{2} + x + 1}} =

[t = \frac{\sqrt{3}}{2} (2 x + 1 + 2 \sqrt{x^{2} + x + 1} \to d x = \frac{\sqrt{3 (x^{2} + x + 1)}}{2 x + 1 + 2 \sqrt{x^{2} + x + 1}}]

= - 16 \int \frac{t}{{(\sqrt{3} t^{2} - 2 t - \sqrt{3})}^{2}} d t =

= \sqrt{3} \int (\frac{1}{2 (\sqrt{3} t - 3)} - \frac{1}{2 (\sqrt{3} t + 1)} - \frac{3}{{(\sqrt{3} t - 3)}^{2}} + \frac{1}{{(\sqrt{3} t + 2)}^{2}}) d t =

= \frac{2 (t + \sqrt{3})}{\sqrt{3} t^{2} - 2 t - \sqrt{3}} - \frac{1}{2} \ln \frac{\sqrt{3} t + 1}{\sqrt{3} t - 3} =

= \frac{\sqrt{x^{2} + x + 1}}{x} + \frac{1}{2} \ln x - \frac{1}{2} \ln (x + 2 + 2 \sqrt{x^{2} + x + 1}) + C

\Rightarrow

\int \frac{\sqrt{x^{2} + x + 2 - \sqrt{4 x^{2} + 4 x + 4}}}{x \sqrt{x^{4} + x^{3} + x^{2}}} d x =

= \frac{- 1 + \sqrt{x^{2} + x + 1}}{x} + \frac{1}{2} \ln x - \frac{1}{2} \ln (x + 2 + 2 \sqrt{x^{2} + x + 1}) + C

Commented by Tawa11 last updated on 06/Sep/21

great sir

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