how to integral the root of 4+9x.^2 and the root of 4x.^2 -9?

∫ sqrt(4+9x^2) dx

Let x = (2/3) tan t
dx = (2/3) sec^2 t dt

4+9x^2 = 4+ 9(4/9) tan^2 t = 4+ 4 tan^2 t = 4(1+tan^2 t) = 4 sec^2 t
sqrt(4+x^2) = 2 sec t

∫ sqrt(4+9x^2) dx = ∫ (2 sec t) (2/3) sec^2 t dt
= (4/3) ∫ sec^3 t dt

Let us integrate ∫ sec^3 t dt

Let I = ∫ sec^3 t dt

Integrate by parts
∫ sec^3 t dt = ∫ sec^2 t sec t dt

dv = sec^2 t ; v = tan t ;
u = sec t ; du = sec t tan t dt

∫ u dv = u v - ∫ v du
I = sec t tan t - ∫ tan t sec t tan t dt
I = sec t tan t - ∫ tan^2 t sec t dt
I = sec t tan t - ∫ (sec^2 t -1) sec t dt
I = sec t tan t - ∫ sec^3 t + ∫ sec t dt
I = sec t tan t - I + ln ( sec t + tan t )
2I = sec t tan t + ln ( sec t + tan t )
I = (1/2) sec t tan t + (1/2) ln ( sec t + tan t )

∫ sec^3 t dt = (1/2) sec t tan t + (1/2) ln ( sec t + tan t )
(4/3) ∫ sec^3 t dt = (2/3) sec t tan t + (2/3) ln ( sec t + tan t )

transform t back to x

The substitution was x = (2/3) tan t
tan t = 3x/2
sec t = sqrt( 1+ tan^2 t) = sqrt( 1+ 9x^2 / 4) = sqrt(4+9x^2) /sqrt(4) = (1/2) sqrt(4+9x^2)

(2/3) sec t tan t + (2/3) ln ( sec t + tan t ) = (2/3)(1/2) sqrt(4+9x^2) (3x/2) + (2/3) ln ( sqrt(4+9x^2) /2 + 3x/2 )
= (1/2) x sqrt(4+9x^2) + (2/3) ln ( sqrt(4+9x^2) /2 + 3x /2 )

∫ sqrt(4+9x^2) dx = (1/2) x sqrt(4+9x^2) + (2/3) ln ( sqrt(4+9x^2) /2 + 3x /2 ) + C
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∫ sqrt(4x^2-9) dx

Let x = (3/2) sec t
dx = (3/2) sec t tan t dt

4x^2-9 = 4(9/4) sec^2 t - 9 = 9 sec^2 t - 9 = 9 tan^2 t
sqrt(4x^2-9) = 3 tan t

∫ sqrt(4x^2-9) dx = ∫ (3 tan t) ( 3/2) sec t tan t dt
= (9/2) ∫ tan^2 t sec t dt
= (9/2) ∫ (sec^2 t - 1) sec t dt
= (9/2) ∫ sec^3 t dt - (9/2) ∫ sec t dt

substitute ∫ sec^3 t dt = (1/2) sec t tan t + (1/2) ln ( sec t + tan t ) ( previous problem)
= (9/2) [ (1/2) sec t tan t + (1/2) ln ( sec t + tan t ) ] - (9/2) ∫ sec t dt
= (9/4) sec t tan t + (9/4) ln (sec t + tan t) - (9/2) ln (sec t + tan t)
= (9/4) sec t tan t - (9/4) ln (sec t + tan t)

The substitution was x= (3/2) sec t
sec t = 2x/3
tan t = sqrt( sec^2 t -1) = sqrt( 4x^2 /9 -1) = (1/3) sqrt(4x^2-9)

(9/4) sec t tan t - (9/4) ln (sec t + tan t) = (9/4) (2x/3) (1/3) sqrt(4x^2-9) - (9/4) ln ( 2x /3 + sqrt(4x^2-9) /3)
= (1/2) x sqrt(4x^2-9) - (9/4) ln ( 2x /3 + sqrt(4x^2-9) /3)

∫ sqrt(4x^2-9) dx = (1/2) x sqrt(4x^2-9) - (9/4) ln ( 2x /3 + sqrt(4x^2-9) /3) + C

I think you mean √(4+9x²) dx and the substitution x=(2/3)tanu gives
the integral of (2/3)sec³u du. Wikipedia (how to find the integral of secant cubed)
gives the required integral to be
(1/3)secutanu+(1/3)ln|secu+tanu| and you convert back to x using tanu=3x/2
with secu=½√(4+9x²) which I leave you to do.

The second can be done using the substitution x=(3/2)secu, dx=(3/2)secu.tanu
This leads to the integral of (9/2)secu.tan²u.du which is the integral of
(9/2)[secu - sec³u] du . The integral of secu is ln|secu + tanu| and that of sec³u is above and you can get the answer having back substituted for x.
It should give ½x√(4x²-9) -(9/4)ln|2x+√(4x²-9)| + C

http://www.symbolab.com/solver/integral-...

hope this link helps u :)

step-by-step solutions for both:

http://symbolab.com/solver/system-of-equations-calculator/%5Cint%5Csqrt%7B(9x%5E%7B2%7D%2B4)%7D/?origin=button


http://symbolab.com/solver/system-of-equations-calculator/%5Cint%20%5Csqrt%7B(4x%5E%7B%5E2%7D-9)%7D/?origin=button

integral of 4+9x^2 is (4+9x)^3/27
integral of 4x^2-9 is 4x^3/3 -9x
always remember that in integration you need to add 1 to power and in differentiation you need to minus 1