what is the integral for ∫x/sqrt(-171-7x^(2)+70x)? explain please. thank you.?

Answer 1

∫x/√(-171-7x^2+70x) dx
= (1/14) [∫((14x - 70)/√(-171-7x^2+70x) dx + ∫70/√(-171-7x^2+70x) dx]
let u = -171-7x^2+70x, du = -14x + 70
= (1/14) ∫(-du/√u) + 5∫dx/√(-171-7x^2+70x)
= (1/14) (-u^(3/2) / (3/2)) + (5/√7)∫dx/√(-171/7 - x^2 + 10x)
= (1/21) (-(-171-7x^2+70x)^(3/2)) + (5/√7)∫dx/√(4/7 - (x-5)^2)

We can handle this with partial fractions, though it gets messy:

Now (4/7 - (x-5)^2) = (2/√7 + x - 5) (2/√7 - x + 5).
So write 1/(4/7 - (x-5)^2) = A/(2/√7+x-5) + B/(2/√7-x+5)
<=> 1 = A(2/√7 - x + 5) + B(2/√7 + x - 5)
from which we get
(constants) 1 = 2/√7(A+B) + 5(A-B)
(x terms) 0 = -A + B
Hence B = A and we get 1 = 4A/√7, hence A = B = √7 / 4.

Thus 1/√(4/7 - (x-5)^2)
= 1/√[(√7/4) (1/(2/√7+x-5) + 1/(2/√7-x+5))]
= 1/(2qrt(7)) (1/√(2/√7+x-5) + 1/√(2/√7-x+5))
where "qrt" represents the fourth root.

So the integral is
(1/21) (-(-171-7x^2+70x)^(3/2)) + (5/√7)∫dx/√(4/7 - (x-5)^2)
= (1/21) (-(-171-7x^2+70x)^(3/2)) + (5/√7)(1/(2qrt(7))) ∫[(2/√7+x-5)^(-1/2) + (2/√7-x+5)^(-1/2)] dx
= (1/21) (-(-171-7x^2+70x)^(3/2)) + (5/√7)(1/(2qrt(7))) [(2/√7+x-5)^(1/2)/(1.(1/2)) + (2/√7-x+5)^(1/2)/(-1.(1/2))] + c
= (1/21) (-(-171-7x^2+70x)^(3/2)) + (5/qrt(7^3)) [√(2/√7+x-5) - √(2/√7-x+5)] + c.

It will no doubt be a good exercise for you to differentiate this expression and confirm that you get back the original integrand. ;-)

Another, equally good exercise would be to verify that my and Dan's answers are the same (using cos x = √(1-sin^2 x), for instance).

Answer 2

- 7 x^2 + 70 x - 171
= - 7 (x^2 - 10x + (171 / 7))
= - 7 (x^2 - 10x + 25 + ((171 / 7) - 25))
= - 7 ((x - 5)^2 + ((171 - (7 * 25)) / 7)
= - 7 ((x - 5)^2 + ((171 - 175) / 7))
= - 7 ((x - 5)^2 - 4/7)
= - 7 (x - 5)^2 + (7 * (4/7))
= - 7 (x - 5)^2 + 4
= 4 - 7 (x - 5)^2

So substitute: sqrt(7) (x - 5) = 2 sin t

The x dx in the numerator become:
x -> 5 + ((2 sin t) / sqrt(7))
dx -> ((2 cos t) / sqrt(7)) dt

The denominator becomes
sqrt(4 - 7 (x - 5)^2) -> sqrt(4 - (2 sin t)^2) = sqrt(4 - 4 (sin t)^2) = sqrt(4 (1 - (sin t)^2)) = sqrt(4 (cos t)^2) = 2 cos t

Altogether you get:

integral of (x dx) / sqrt(- 171 - 7 x^2 + 70 x)
= integral of (5 + ((2 sin t) / sqrt(7)) (((2 cos t) / sqrt(7)) dt) / (2 cos t)
= integral of (1/7) (5 sqrt(7) + 2 sin t) dt
= (1/7) (5 sqrt(7) t - 2 cos t) + C

t = inverse sin (sqrt(7) (x - 5) / 2)
2 cos t = the original denominator

So the integral is:

(1/7) ((5 sqrt(7) inverse sin (sqrt(7) (x - 5) / 2)) - sqrt(- 171 - 7 x^2 + 70 x)) + C

As always, you can verify the result by taking the derivative and showing that it equals x / sqrt(- 171 - 7 x^2 + 70 x). :-)

Dan