x-2-1-x-2-dx-

Question Number 85813 by jagoll last updated on 25/Mar/20

\int x^{2} \sqrt{1 + x^{2}} dx ?

Commented by jagoll last updated on 25/Mar/20

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

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

= \frac{1}{3} x {(1 + x^{2})}^{3 / 2} - \frac{1}{3} \int {(1 + x^{2})}^{3 / 2} dx

let J = \int {(1 + x^{2})}^{3 / 2} dx

[x = \tan t]

J = \int \sec^{5} t dt

= \sec^{3} t \tan t - \int 3 \sec^{5} t dt + \int 3 \sec^{3} t dt

4 J = \sec^{3} t \tan t + 3 (\frac{1}{2} \sec t \tan t - \frac{1}{2} \ln ∣ \sec t + \tan t ∣)

J = \frac{1}{4} \sec^{3} t \tan t + \frac{3}{8} \sec t \tan t -

\frac{3}{8} \ln ∣ \sec t + \tan t ∣

Answered by john santu last updated on 25/Mar/20

l e t x = \sinh t \Rightarrow d x = \cosh t d t

I = \int \sinh^{2} t \cosh t (\cosh t d t)

= \int \sinh^{2} t \cosh^{2} t d t

= \frac{1}{4} \int \sinh^{2} 2 t d t

= \frac{1}{8} \int (\cosh 4 t - 1) d t

= \frac{1}{8} (\frac{\sinh 4 t}{4} - t) + c

= \frac{1}{8} [(\sinh t \cosh t) (1 + 2 \sinh^{2} t) - t] + c

= \frac{1}{8} [(x \sqrt{1 + x^{2}}) (1 + 2 x^{2}) - \sinh^{- 1} x] + c

Commented by jagoll last updated on 25/Mar/20

waw thank you sir . i try via integration

by parts

Answered by MJS last updated on 25/Mar/20

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

[x = \sinh \ln t = \frac{t^{2} - 1}{2 t} \Leftrightarrow t = x + \sqrt{x^{2} + 1} \to d x = \frac{t^{2} + 1}{2 t^{2}}]

= \frac{1}{16} \int t^{3} d t - \frac{1}{8} \int \frac{d t}{t} + \frac{1}{16} \int \frac{d t}{t^{5}} =

= \frac{1}{64} t^{4} - \frac{1}{8} \ln t - \frac{1}{64 t^{4}} = \frac{t^{8} - 1}{64 t^{4}} - \frac{1}{8} \ln t =

= \frac{1}{8} (x (2 x^{2} + 1) \sqrt{x^{2} + 1} - \ln (x + \sqrt{x^{2} + 1})) + C

Commented by jagoll last updated on 25/Mar/20

what formula \sinh (\ln x) sir ?

Commented by MJS last updated on 25/Mar/20

{it}^{'} s just as it is

I had the idea to use this when I had to

solve integrals in 2 steps, 1^{st} a hyperbolic

substitution, 2^{nd} get rid of e^{t} . you can do

it in one step

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