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# How do you integrate ((x^2-9)^(1/2)/x dx x>3?

The answer is (x^2-9)^(1/2)-3sec^(-1)(x/3)+C

I am unsure how to solve this. Can anyone help?

### 3 Answers

- PseudonymLv 59 years agoFavorite Answer
When faced with an integration problem which involves square roots (or any radicals, for that matter), shift it into the denominator. This helps enormously.

∫ √(x²-9) / x dx

= ∫ (x²-9) / x √(x²-9) dx

This doesn't look like much of an improvement, but consider this:

d/dx (√f(x)) = f'(x)/2√f(x)

Differentiation moves square roots from the numerator into the denominator. So integration should do the opposite.

There are several ways to attack this problem. They all boil down to partial fractions in one way or another. If you recognise this standard integral:

d/dx Arcsec[x] = 1/x√(x²-1)

(I know that's not in the usual form of an integral table, but bear with me) you can recover:

d/dx Arcsec[x/3] = 3/x√(x²-9)

So it looks like the integral probably has this form:

∫ (x²-9) / x √(x²-9) dx = A √(x²-9) + B Arcsec[x/3] + C

for some constants A and B. To find the constants, differentiate both sides:

(x²-9) / x √(x²-9) = Ax/√(x²-9) + 3B / x√(x²-9)

Cross-multiply:

x²-9 = Ax² + 3B

which has the trivial solution:

A=1

B=-3

I hope you can see that this is just a generalisation of partial fractions.

- Anonymous9 years ago
The difference of squares with the variable being first screams for a trig sub; however, your goal in constructing the substution means we want the coefficients and constant terms to be unity so we can use the Pythagorean Theorem for trig functions (it's the heart of the substitution). For this problem the we need to eliminate the 9, so we choose 3 since 3^2 = 9. Then the substitution becomes

x = 3 sec t and dx = 3 sec t tan t dt

Plugging these into the integrand yields

sqrt( x^2 - 9) / x dx = sqrt[ 9 sec^2 t - 9] / [3 sec t] 3 sec t tan t dt

= sqrt[ 9 (sec^2 -1)] tan t dt

= sqrt(9) sqrt( tan^2 t) tan t dt

= 3 tan t * tan t dt

= 3 tan^2 t dt

= 3 (sec^2 t - 1) dt

Using the above then as my integrand I have

3 int(sec^2 t dt) - 3 int( dt) = 3 tan t - 3 t + C

Using the original substitution you have x/3 = sec t we have plugging back in using a right triangle trig

3 * sqrt(x^2 - 9)/3 - 3 acrsec(x/3) + C = sqrt(x^2 - 9) - 3 arcsec(x/3) + C

which is the desired result.

- ?Lv 44 years ago
u = x^2 + a million u - a million = x^2 sqrt(u - a million) = |x| If we take an indispensable from x = a million to x = 2, or certainly as you suggested, from x = 2 to x = a million, in the two case x is often valuable so we are in a position to drop easily the fee operation. i visit think of which you meant a million to be the decrease sure. If not, merely exchange the connect up our very final answer. x = sqrt(u - a million) dx = a million /(2*sqrt(u - a million)) du Now we are in a position to rewrite our indispensable as follows, changing our limits of integration to be in terms of u particularly of x ?[from u=2 to u=5] (sqrt(u - a million)*u^3) / (2sqrt(u - a million)) du = ?[2:5] (u^3)/2 du = (a million/8)u^4 ]{2:5} we are in a position to now enter immediately = (a million/8)[5^4 - 2^4] = 609/8 or return to a kind in terms of x =(a million/8)(x^2 + a million)^4]{a million:2} =(a million/8)[5^4 - 2^4] = 609/8 --charlie