# Find the arc length of the curve y=ln(x) from x=1 to x=2 I got as far as `intsqrt(x^2+1)/x` Now, since this is in the section of trigonometric substitutions I went further on I said...

Find the arc length of the curve y=ln(x) from x=1 to x=2

I got as far as `intsqrt(x^2+1)/x`

Now, since this is in the section of trigonometric substitutions I went further on I said x=(1)tan`theta` and `dx=sec^2(theta)d theta`

and when I plugged it in I got: `int_1^2(sec^3(theta))/tanthetad theta`

Now, I'm stuck and don't know what to do.

*print*Print*list*Cite

### 1 Answer

It looks like you set up the original integral correctly. You used the trigonometric substitution correctly also (except, remember to change the integration limits to arctan(1) and arctan(2)) and it could be done this way, but this is not the easiest way to take this integral. The substitution that I believe is effective here is

`u^2= x^2 + 1` (u then goes from 2 to 5)

2udu = 2xdx

Plugging this into the integral results in

`int _2 ^5 sqrt(u^2)/x * (udu)/x = int_2 ^5 u^2/x^2 du = int_2 ^5 u^2/(u^2 - 1) du`

The expression into the integral can then be broken up into partial fractions:

`u^2/(u^2 - 1) = (u^2 - 1 + 1)/(u^2 - 1) = 1 + 1/(u^2 - 1) = 1 + 1/2(1/(u-1) - 1/(u + 1))`

Each of the three parts of this expression now can be integrated. The antiderivative of 1 is u, the antiderivative of `1/(u-1) ` is ln(u-1) and the antiderivative of `1/(u+1)` is ln(u+1). So,

`int_2 ^5 u^2/(u^2 - 1) = [u + 1/2 ln(u-1) - 1/2ln(u+1)] |_2 ^ 5`

After plugging in the limits, we get

5 + (1/2)ln(4) - (1/2)ln(6) - 2 - (1/2)ln(1) + (1/2)ln(3) = 3+ (1/2)ln(2) (this results from combining the logarithms by using the rule for logs of product/quotient.)

**So the length of arc will be**

`3 + 1/2ln(2)`