x-2-1-x-1-2x-3-dx-

Question Number 94609 by M±th+et+s last updated on 20/May/20

\int \frac{x^{2} - 1}{\sqrt{x + 1} + \sqrt{2 x + 3}} d x

Answered by mathmax by abdo last updated on 20/May/20

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

= \int \frac{x^{2} - 1}{x + 2} \sqrt{2 x + 3} d x - \int \frac{x^{2} - 1}{x + 2} \sqrt{x + 1} dx = A - B changement

\sqrt{2 x + 3} = t give 2 x + 3 = t^{2} \Rightarrow x = \frac{t^{2} - 3}{2} \Rightarrow

A = \int \frac{\frac{1}{4} {(t^{2} - 2)}^{2} - 1}{\frac{t^{2} - 3}{2} + 2} \times t (t) dt = \frac{1}{2} \int \frac{{(t^{2} - 2)}^{2} - 4}{t^{2} - 3 + 4} t^{2} dt

= \frac{1}{2} \int \frac{t^{2} (t^{4} - {4 t}^{2})}{t^{2} + 1} dt = \frac{1}{2} \int \frac{t^{6} - {4 t}^{4}}{t^{2} + 1} dt = \frac{1}{2} \int \frac{t^{4} (t^{2} + 1) - {5 t}^{4}}{t^{2} + 1} dt

= \frac{1}{2} \int t^{4} dt - \frac{5}{2} \int \frac{t^{4} - 1 + 1}{t^{2} + 1} dt = \frac{t^{5}}{10} - \frac{5}{2} \int (t^{2} - 1) dt - \frac{5}{2} arctant

= \frac{t^{5}}{10} - \frac{5}{6} t^{3} + \frac{5}{2} t - \frac{5}{2} \arctan (t) + c_{0}

= \frac{{(\sqrt{2 x + 3})}^{5}}{10} - \frac{5}{6} {(\sqrt{2 x + 3})}^{3} + \frac{5}{2} (\sqrt{2 x + 3}) - \frac{5}{2} \arctan (\sqrt{2 x + 3}) + c_{0}

changement \sqrt{x + 1} = t give x = t^{2} - 1 \Rightarrow

B = \int \frac{{(t^{2} - 1)}^{2} - 1}{t^{2} - 1 + 2} \times t (2 t) dt = 2 \int \frac{t^{2} (t^{4} - {2 t}^{2})}{t^{2} + 1} dt

= 2 \int \frac{t^{6} - {2 t}^{4}}{t^{2} + 1} dt = 2 \int \frac{t^{4} (t^{2} + 1) - {3 t}^{4}}{t^{2} + 1} dt

= 2 \int t^{4} dt - 6 \int \frac{t^{4}}{t^{2} + 1} dt = \frac{2}{5} t^{5} - 6 \int \frac{t^{4} - 1 + 1}{t^{2} + 1} dt

= \frac{2}{5} t^{5} - 6 \int (t^{2} - 1) dt - 6 \arctan (t)

= \frac{2}{5} t^{5} - {2 t}^{3} + 6 t - 6 \arctan (t) + c_{1}

= \frac{2}{5} {(\sqrt{x + 1})}^{5} - 2 {(\sqrt{x + 1})}^{3} + 6 \sqrt{x + 1} - 6 \arctan (\sqrt{x + 1}) + c_{1}

I = A - B

Commented by M±th+et+s last updated on 20/May/20

w e l l d o n e .

t h a n k y o u s i r

Commented by mathmax by abdo last updated on 20/May/20

you are welcome

Answered by MJS last updated on 20/May/20

I found this; integration is easy but inserting

at the end is tricky \dots

\int \frac{x^{2} - 1}{\sqrt{x + 1} + \sqrt{2 x + 3}} d x =

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

= \frac{\sqrt{2}}{128} \int \frac{{(t - 1)}^{3} {(t + 1)}^{3} (t^{2} + 1) (t^{4} - 18 t^{2} + 1)}{t^{6} ((1 + \sqrt{2}) t^{2} - 1 + \sqrt{2})} d t =

= Σ of the following :

(1) - 12 (1 - \sqrt{2}) \int \frac{d t}{t^{2} + 3 - 2 \sqrt{2}} = 12 \arctan ((1 + \sqrt{2}) t)

(2) \frac{2 - \sqrt{2}}{128} \int t^{4} d t = \frac{2 - \sqrt{2}}{640} t^{5}

(3) - \frac{50 - 27 \sqrt{2}}{128} \int t^{2} d t = - \frac{50 - 27 \sqrt{2}}{384} t^{3}

(4) \frac{166 - 109 \sqrt{2}}{64} \int d t = \frac{166 - 109 \sqrt{2}}{64} t

(5) - \frac{166 + 109 \sqrt{2}}{64} \int \frac{d t}{t^{2}} = \frac{166 + 109 \sqrt{2}}{64 t}

(6) \frac{50 + 27 \sqrt{2}}{128} \int \frac{d t}{t^{4}} = - \frac{50 + 27 \sqrt{2}}{384 t^{3}}

(7) - \frac{2 + \sqrt{2}}{128} \int \frac{d t}{t^{6}} = \frac{2 + \sqrt{2}}{640 t^{5}}

=

12 \arctan ((1 + \sqrt{2}) (\sqrt{2 (x + 1)} + \sqrt{2 x + 3})) +

- \frac{2}{5} (x^{2} - 3 x + 11) \sqrt{x + 1} +

+ \frac{1}{15} (6 x^{2} - 17 x + 51) \sqrt{2 x + 3} + C

Commented by M±th+et+s last updated on 20/May/20

t h a n k y o u s i r . n i c e s o r k