Question

plz find the answer as soon as possible​

Answer 1

To find--->

-------------

dy x

∫ ------ = ∫ ----------------- dx

y √(x² - 4 )

Solution--->

---------------

1 x

=> ∫ ------- dy = ∫ -------------- dx + c

y √(x² - 4 )

x

=> log y = ∫ --------------- dx +c

√(x² - 4)

Let x² - 4 = t

Differentiating with respect to x

( 2x - 0 )dx = dt

x dx = dt/2

1

=> log y = ∫ --------------- (dt/2) + c

√t

= 1/2 ∫ t⁻¹/² dt + c

We have a formula

xⁿ⁺¹

∫ xⁿ dx = ----------- + c applying it here

n + 1

t⁻¹/² ⁺ ¹

= 1/2 ( ----------- ) + c

-1/2 +1

t¹/²

= 1/2 ( -----------) + c

1/2

= t¹/² + c

Putting value of t

=> log y = √ (x² - 4 ) + c

Additional information--->

-------------------------------------

1) ∫ 1/x dx= logx + c

2) ∫eˣ dx = eˣ + c

3) ∫ aˣ dx = aˣ / loga + c

4) ∫ Sinx dx = - Cosx + c

5) ∫ Cosx dx = Sinx + c

6) ∫ Sec²x dx = tanx + c

7) ∫ Secx tanx dx = Secx + c

8) ∫ Cose²x dx = - Cotx + c

9) ∫ Cosecx Cotx dx = - Cosecx + c