Can anyone help to compute 3 integrals?

Question

I tried to work on them, but I can't solve with either trigonometric u substitution or partial fraction.

The problems require to using the following integral formulas:
#int(du)/sqrt(a^2-u^2)=sin^-1(u/a)#
#int(du)/sqrt(u^2+a^2)=ln(u+sqrt(u^2+a^2))# #int(du)/(u^2+a^2)=1/atan^-1(u/a)#

1.#intdx/sqrt(x^2+6x+10#

2.#int(dx)/sqrt(6x-x^2)#

3.#int(dx)/(2x^2+6x+5)#

Paisley Johnson · Accepted Answer

Answers below...

Initial integral: Completing the square on #x^2 + 6x+10#: #x^2 +6x+10 -= (x+3)^2+1 # #=> int (dx)/(sqrt( x^2 +6x+10) ) -= int (dx)/(sqrt((x+3)^2+1) # Let #u = x+3 # #=> du = dx # #=> int (du)/(sqrt(u^2+1)) # # = ln(u+sqrt(u^2+1) ) + c " Using the formulae you stated! " # Subtituting # u = x+3 # # = ln( x+3 + sqrt( (x+3)^2 + 1 ) ) + c # #color(red)( = ln(x+3+sqrt( x^2 +6x+10) )+c # The second integral We can use the same concept here. #6x-x^2 -= 9-(x-3)^2 # #=> int (dx)/sqrt(9-(x-3)^2 ) # #u = x-3 # #du = dx# #=> int (du)/sqrt((9-u^2) # # = sin^(-1) ( u /3 ) + c # As #a^2 = 9 => a = 3 # in this case... Substituting #u = x-3 # #color(red)(= sin^(-1) (( x-3)/3 ) + c # The third integral Applying the same concept: #2x^2+6x+5 -= 2(x+3/2)^2 + 1/2 # #int (dx)/( 2(x+3/2)^2 + 1/2 ) # Let # u = sqrt(2) ( x+3/2) # #u^2 = 2(x+3/2)^2 # #du = sqrt(2) dx # #=>sqrt(2)/2 du = dx # #=> sqrt(2)/2int (du)/(u^2 + 1/2) # For this case #a^2 = 1/2 => a = sqrt(2)/2 # #=> = sqrt(2)/2 * sqrt(2)tan^(-1) ( sqrt(2)*u )+c # # = tan^(-1)(sqrt(2)u) + c # Substituting # u = sqrt(2)(x+3/2) # #color(red)( = tan^(-1) ( 2x+3 ) + c #

Luke Alvarez · Answer

#1. ln|(x+3)+sqrt(x^2+6x+10)|+C_1#.

#2. sin^-1((x-3)/3)+C_2#.

#3. tan^-1(2x+3)+C_3#.

1. #intdx/sqrt(x^2+6x+10)=I_1," say."# #:. I_1=intdx/sqrt{(x+3)^2+1}#. Letting, #(x+3)=u, dx=du#. #:. I_1=int(du)/sqrt(u^2+1^2)#, #=ln|(u+sqrt(u^2+1^2))|#, #rArr I_1=ln|(x+3)+sqrt(x^2+6x+10)|+C_1#. 2. #intdx/sqrt(6x-x^2)=I_2," say."# #:. I_2=intdx/sqrt(9-9+6x-x^2)=intdx/sqrt{3^2-(x-3)^2}#. We subst. #x-3=y," so that, "dx=dy#. #:. I_2=intdy/sqrt(3^2-y^2)#, #=sin^-1(u/3)#. #rArr I_2=sin^-1((x-3)/3)+C_2#. 3. #intdx/(2x^2+6x+5)=I_3," say"#. #:. I_3=intdx/{2(x^2+3x+5/2)#, #=1/2intdx/(x^2+3x+9/4+1/4)#, #=1/2intdx/{(x+3/2)^2+(1/2)^2#. Taking #(x+3/2)=t," we have, "dx=dt#. #:. I_3=1/2intdt/(t^2+(1/2)^2#, #=1/2{1/(1/2)tan^-1(t/(1/2))}#, #=tan^-1(2t)#. #rArr I_3=tan^-1(2(x+3/2)=tan^-1(2x+3)+C_3#. Spread the joy and have fun with math!

Ezra Adams · Answer

undefined Of course, I'm here to help. Please provide the integrals you need computed, and I'll assist you with their solutions.