what is the integral of 16 + 6x − x2 dx?

look at the attachment. its the sqrt

∫ sqrt (16 +6x-x^2) dx

16+6x-x^2 = -(x^2-6x-16) = -( x^2-6x+9-9-16) = - ( (x-3)^2 -25) = sqrt( 25- (x-3)^2 )
∫ sqrt (16 +6x-x^2) dx = ∫ sqrt( 25- (x-3)^2 ) dx

To integrate ∫ sqrt( 25- (x-3)^2 ) dx , let u=x-3
du = dx
∫ sqrt( 25- (x-3)^2 ) dx = ∫ sqrt(25-u^2) du

Let u= 5 sin t
du = 5 cos t dt
25-u^2 = 25 - 25 sin^2 t = 25(1-sin^2 t) = 25 cos^2 t
sqrt(25-u^2) = 5 cos t

∫ sqrt(25-u^2) du = ∫ (5 cos t) ( 5 cos t) dt
= 25 ∫ cos^2 t dt

Let us integrate ∫ cos^2 t dt

∫ cos^2 t dt = ∫ (1+cos 2t) dt / 2
= ∫ (1/2) dt + ∫ (1/2) cos 2t dt
= t / 2 + (1/2) ∫ cos 2t dt --------------------- (1)

let s = 2t
ds = 2 dt
dt = (1/2) ds
(1/2) ∫ cos 2t dt = (1/2)(1/2) ∫ cos s ds
(1/2) ∫ cos 2t dt = (1/4) ∫ cos s ds
(1/2) ∫ cos 2t dt = (1/4) sin s
(1/2) ∫ cos 2t dt = (1/4) sin 2t
= (1/4) ( 2 sin t cos t )
= (1/2) sin t cos t

substitute this into (1)
∫ cos^2 t dt = (1/2) t + (1/2) sin t cos t
25 ∫ cos^2 t dt = (25/2) t + (25/2) sin t cos t

our substitution was u= 5 sin t
sin t = u/5
t = sin^-1(u/5)
cos t = sqrt(1-sin^2 t) = sqrt( 1- u^2/25) = (1/5) sqrt(25-u^2)

(25/2) t + (25/2) sin t cos t = (25/2) sin^-1(u/5) +(25/2) (u/5) (1/5) sqrt(25-u^2)
= (25/2) sin^-1(u/5) +(1/2) u sqrt(25-u^2)

replace u by (x-3)
= (25/2) sin^-1((x-3)/5) +(1/2) (x-3) sqrt(25-(x-3)^2)
= (25/2) sin^-1((x-3)/5) +(1/2) (x-3) sqrt(16+6x-x^2)

∫ sqrt (16 +6x-x^2) dx = (25/2) sin^-1((x-3)/5) +(1/2) (x-3) sqrt(16+6x-x^2) + C
-------------- --------------- ----------------- ---------------- -------

∫ sqrt(x^2+6x) dx

x^2+6x = x^2+6x+9-9 = sqrt( (x+3)^2 -9)
∫ sqrt(x^2+6x) dx = ∫ sqrt ((x+3)^2 -9)

Let u= x+3
du = dx
∫ sqrt ((x+3)^2 -9) = ∫ sqrt(u^2-9) du

Let u = 3 sec t
du = 3 sec t tan t dt
u^2-9 = 9 sec^2 t - 9 = 9(sec^2 t -1) = 9 tan^2 t
sqrt(u^2-9) = 3 tan t

∫ sqrt(u^2-9) du = ∫ (3 tan t) ( 3 sec t tan t ) dt
= 9 ∫ sec t tan^2 t dt
= 9 ∫ sec t (sec^2 t -1) dt
= 9 ∫ sec^3 t dt - 9 ∫ sec t dt
= 9 ∫ sec^3 t dt - 9 ln (sec t + tan t) --------------- (1)

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 )
9 ∫ sec^3 t dt = (9/2) sec t tan t + (9/2) ln ( sec t + tan t )

substitute this into (1)
9 ∫ sec^3 t dt - 9 ln (sec t + tan t) = (9/2) sec t tan t + (9/2) ln ( sec t + tan t ) - 9 ln (sec t + tan t)
= (9/2) sec t tan t - (9/2) ln ( sec t + tan t )

transform t back to u
our substitution was u = 3 sec t
sec t = u/3
tan t = sqrt( sec^2 t -1) = sqrt( u^2/9 -1) = (1/3) sqrt(u^2-9)

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

replace u by x+3
= (1/2) (x+3) sqrt((x+3)^2-9) - (9/2) ln ( (x+3) /3 + sqrt((x+3)^2-9) / 3)
= (1/2) (x+3) sqrt( x^2+6x) - (9/2) ln ( (x+3) /3 + sqrt(x^2+6x) / 3)

∫ sqrt(x^2+6x) dx = (1/2) (x+3) sqrt( x^2+6x) - (9/2) ln ( (x+3) /3 + sqrt(x^2+6x) / 3) + C
-------------------- ------------ ------------
∫ x^2 dx / (15+4x-4x^2)^(3/2)

15+4x-4x^2 = -(4x^2-4x-15) = -( (2x)^2 - 2(2x)(1) + 1 - 1 -15) = - ( (2x-1)^2 -16) = 16- (2x-1)^2
(15-4x-4x^2)^(3/2) = ( 16 - (2x-1)^2 )^(3/2)

∫ x^2 dx / (15+4x-4x^2)^(3/2) = ∫ x^2 dx / ( 16 -(2x-1)^2)^(3/2)

Let u= 2x-1
du = 2 dx
dx = (1/2) du
x= (u+1)/2

∫ x^2 dx / ( 16 -(2x-1)^2)^(3/2) = (1/8) ∫ (u+1)^2 / (16-u^2)^(3/2) du

Let u= 4 sin t
du = 4 cos t dt
16-u^2 = 16-16sin^2 t = 16 cos^2 t
(16-u^2)^(3/2) = (16 cos^2 t)^(3/2) = 64 cos^3 t

(1/8) ∫ (u+1)^2 / (16-u^2)^(3/2) du = (1/8) ∫ (4 sin t +1)^2 (4 cos t ) dt / cos^3 t
= (1/2)(1/64) ∫ (4 sin t +1)^2 / cos^2 t dt
= (1/128) ∫ (4 sin t +1)^2 sec^2 t dt
= (16/128) ∫ sin^2 t sec^2 t dt + (8/128) ∫ sin t sec^2 t dt + (1/128) ∫ sec^2 t dt
= (1/8) ∫ sin^2 t sec^2 t dt + (1/16) ∫ sin t sec^2 t dt + (1/128) ∫ sec^2 t dt
= (1/8) ∫ tan^2 t dt + (1/16) ∫ tan t sec t dt + (1/128) ∫ sec^2 t dt
= (1/8) ∫ (sec^2 t -1) dt + (1/16) ∫ tan t sec t dt + (1/128) ∫ sec^2 t dt
= (1/8) ∫ sec^2 t dt - (1/8) ∫ dt + (1/16) ∫ tan t sec t dt + (1/128) ∫ sec^2 t dt
= (17/128) ∫ sec^2 t dt - (1/8) ∫ dt + (1/16) ∫ tan t sec t dt
= (17/128) tan t - (1/8) t + (1/16) sec t

Transform t back to u
our substitution was u = 4 sin t
sin t = u/4
t = sin^-1(u/4)
tan t = sin t /cos t = sin t / sqrt(1-sin^2 t) = (u/4) / sqrt(1-u^2/16) = u / sqrt(16-u^2)
sec t = sqrt(1+tan^2 t) = sqrt( 1+ u^2 /(16-u^2) = 4 / sqrt(16-u^2)

(17/128) tan t - (1/8) t + (1/16) sec t = 17u / 128sqrt(16-u^2) - (1/8) sin^-1(u/4) + 1 / 4sqrt(16-u^2)
replace u by 2x-1
= 17(2x-1) / 128sqrt(16-(2x-1)^2) - (1/8) sin^-1((2x-1)/4) + 1 / 4sqrt(16-(2x-1)^2)

∫ x^2 dx / (15+4x-4x^2)^(3/2) =
17(2x-1) / [128sqrt(16-(2x-1)^2)] - (1/8) sin^-1((2x-1)/4) + 1 / [4sqrt(16-(2x-1)^2)]+ C
=
17(2x-1) / [128sqrt(15+4x-4x^2)] - (1/8) sin^-1((2x-1)/4) + 1 / [4sqrt(15+4x-4x^2)]+ C