1-x-x-dx-

Question Number 12377 by tawa last updated on 20/Apr/17

\int \frac{\sqrt{1 + \sqrt{x}}}{x} dx

Answered by mrW1 last updated on 21/Apr/17

t^{2} = \sqrt{x}

t^{4} = x

4 t^{3} d t = d x

I = \int \frac{\sqrt{1 + \sqrt{x}}}{x} d x = \int \frac{\sqrt{1 + t^{2}}}{t^{4}} \times 4 t^{3} d t

= 4 \int \frac{\sqrt{1 + t^{2}}}{t} d t = 2 \int \frac{\sqrt{1 + t^{2}}}{t^{2}} d (t^{2})

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

\int \frac{\sqrt{1 + s}}{s} d s = 2 \sqrt{1 + s} + \ln \frac{\sqrt{1 + s} - 1}{\sqrt{1 + s} + 1} + C

I = 4 \sqrt{1 + s} + 2 \ln \frac{\sqrt{1 + s} - 1}{\sqrt{1 + s} + 1} + C

= 4 \sqrt{1 + \sqrt{x}} + 2 \ln \frac{\sqrt{1 + \sqrt{x}} - 1}{\sqrt{1 + \sqrt{x}} + 1} + C

Commented by tawa last updated on 21/Apr/17

wow . God bless you sir .

Answered by ajfour last updated on 21/Apr/17

l e t x = \tan^{4} t \Rightarrow d x = 4 \tan^{3} t \sec^{2} t

\int \frac{\sqrt{1 + \sqrt{x}}}{x} d x = \int \frac{(\sec t) (4 \tan^{3} t \sec^{2} t)}{\tan^{4} t} d t

= 4 \int cosec t \sec^{2} t d t

= 4 [cosec t \tan t - \int \tan t (- cosec t \cot t) d t]

= 4 [\frac{1}{\cos t} + \int cosec t d t] + c

= 4 \sqrt{1 + \sqrt{x}} + 4 \ln ∣ \sqrt{1 + \frac{1}{\sqrt{x}}} - \sqrt{\frac{1}{\sqrt{x}}} ∣ + c^{'}

= 4 \sqrt{1 + \sqrt{x}} + 4 \ln ∣ \sqrt{1 + \sqrt{x}} - 1 ∣ - \ln ∣ x ∣ + c^{'} .

Commented by tawa last updated on 21/Apr/17

God bless you sir .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Apr/17

x = {c o s}^{2} θ \Rightarrow d x = - 2 s i n θ c o s θ d θ

I = \int \frac{\sqrt{1 + c o s θ}}{{c o s}^{2} θ} d θ = \sqrt{2} \int \frac{c o s \frac{θ}{2}}{{c o s}^{2} θ} (- 2 s i n θ c o s θ) d θ

= - 4 \sqrt{2} \int \frac{{c o s}^{2} \frac{θ}{2} s i n \frac{θ}{2}}{2 {c o s}^{2} \frac{θ}{2} - 1} d θ = 8 \sqrt{2} \int \frac{φ^{2} d φ}{2 φ^{2} - 1} =

4 \sqrt{2} \int \frac{(2 φ^{2} - 1) + 1}{2 φ^{2} - 1} d φ = 4 \sqrt{2} \int (1 - \frac{1}{2 φ^{2} - 1}) d φ =

= 4 \sqrt{2} [φ - \frac{1}{2} \int \frac{((\sqrt{2} φ - 1) - (\sqrt{2} φ + 1))}{(\sqrt{2} φ - 1) (\sqrt{2} φ + 1)} d φ] =

= 4 \sqrt{2} [φ - \frac{1}{2} \int (\frac{1}{\sqrt{2} φ + 1} - \frac{1}{\sqrt{2} φ - 1}) d φ] =

= 4 \sqrt{2} [φ - \frac{1}{2 \sqrt{2}} l n \frac{\sqrt{2} φ + 1}{\sqrt{2} φ - 1}] + C =

= 4 \sqrt{2} c o s \frac{θ}{2} - 2 l n \frac{\sqrt{2} c o s \frac{θ}{2} + 1}{\sqrt{2} c o s \frac{θ}{2} - 1} + C

{c o s}^{2} θ = x \Rightarrow c o s θ = \sqrt{x} \Rightarrow 2 {c o s}^{2} \frac{θ}{2} = 1 + \sqrt{x}

c o s \frac{θ}{2} = \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{2}}

I = 4 \sqrt{2} \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{2}} - 2 l n \frac{\sqrt{2} \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{2}} + 1}{\sqrt{2} \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{2}} - 1} + C =

I = 4 \sqrt{1 + \sqrt{x}} - 2 l n (\frac{\sqrt{1 + \sqrt{x}} + 1}{\sqrt{1 + \sqrt{x}} - 1}) + C .

Commented by tawa last updated on 21/Apr/17

i really appreciate sir .

Commented by tawa last updated on 21/Apr/17

God bless you sir .

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