# Substitution Method

1. Jul 20, 2005

Hi,

I have another problem about substitution Method. I think this method is used to make the problem to solve in easy way but it is making my procedure too long for this problem. Can you solve it by substitution method.

S (x+5)½/x-4 dx

where S is the sign of integral. The answer of this problem is

2(x+5)½+3ln[(x+5)½-3/(x+5)½+3]+c

2. Jul 20, 2005

### Cyrus

You will have to do a trig substitution using either sin or cos. I hope your at least attemping these homework problems before you ask for the anwser, because they seem like homework.

3. Jul 21, 2005

### unggio

i set $$u^2= x+5$$

i get integral:
$$2\int \frac {\2{u^2}}{(u+3)(u-3)} du$$

then do the long division, do partial fractions then i get the same answer as u posted.

answer is : $$2u -3\ln(u+3) + 3\ln(u-3) + c$$

i was wondering how do u know if
$$u=? +\sqrt{x+5} \ or \ u=?-\sqrt{x+5}$$
????

Last edited: Jul 21, 2005
4. Jul 21, 2005

### Cyrus

nice choice of subsitution unggio, i take back having to use trig now that I see what you wrote.

You can learn to use the $$LaTeX$$ in the physics forum.

5. Jul 21, 2005

### Orion1

$$u^2 = x + 5$$

$$2 \int \frac{u^2}{(u + 3)(u - 3)} du = 2 \left( u + \frac{3}{2} \ln [u - 3] - \frac{3}{2} \ln [u + 3] \right) + C$$

$$2 \left( u + \frac{3}{2} \ln [u - 3] - \frac{3}{2} \ln [u + 3] \right) + C= 2u + 3 \ln (u - 3) - 3 \ln (u + 3) + C$$

$$2u + 3 \ln (u - 3) - 3 \ln (u + 3) = 2 \sqrt{x + 5} + 3 \ln (\sqrt{x+5} - 3) - 3 \ln (\sqrt{x + 5} + 3) + C$$

$$\boxed{ \int \frac{\sqrt{x + 5}}{x - 4} dx = 2 \sqrt{x + 5} + 3 \ln (\sqrt{x+5} - 3) - 3 \ln (\sqrt{x + 5} + 3) + C}$$

My research also located an Identity:
$$\tanh^{-1} \left( \frac{u}{3} \right) = \frac{1}{2} ( \ln [u + 3] - \ln [u - 3] )$$

Last edited: Jul 21, 2005
6. Jul 21, 2005

### unggio

how do u decide that u is the positive or negative?

$$u=? +\sqrt{x+5} \ or \ u=?-\sqrt{x+5}$$

7. Jul 21, 2005

### Orion1

The functional identity:
$$\tanh^{-1} \left( \frac{u}{3} \right) = \frac{1}{2} ( \ln [u + 3] - \ln [u - 3] )$$

There are two possible solutions in Quadrants I and III, therefore $$u = \pm \sqrt{x + 5}$$

However, the solution:
$$2 \sqrt{x + 5} + 3 \ln (\sqrt{x+5} - 3) - 3 \ln (\sqrt{x + 5} + 3) + C = 0$$

Has x-intercept only at $$x = 7.953$$, therefore $$u = + \sqrt{x + 5}$$

Last edited: Jul 21, 2005