Question

What is the arc length of `f(x) = x^2-ln(x^2)` on `x in [1,3]`?

Answer 1

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-ln((4sqrt265-55)/45)` units.

Explanation:

`f(x)=x^2-ln(x^2)=x^2-2lnx`

`f'(x)=2x-2/x`

Arc length is given by:

`L=int_1^3sqrt(1+(2x-2/x)^2)dx`

Expand and simplify:

`L=int_1^3sqrt(4x^4-7x^2+4)/xdx`

For ease of algebra, apply the substitution `x^2=u`:

`L=1/2int_1^9sqrt(4u^2-7u+4)/udu`

Multiply numerator and denominator by `sqrt(4u^2-7u+4)`

`L=1/2int_1^9(4u^2-7u+4)/(usqrt(4u^2-7u+4))du`

Rearrange so that the numerator contains the derivative of the denominator:

`L=1/4int_1^9((8u-7)-(7-8/u))/sqrt(4u^2-7u+4)du`

Integration is distributive:

`L=1/4int_1^9(8u-7)/sqrt(4u^2-7u+4)du-1/4int_1^9(7-8/u)/sqrt(4u^2-7u+4)du`

Complete the square in the square root:

`L=1/2[sqrt(4u^2-7u+4)]_ 1^9-int_1^9(7-8/u)/sqrt((8u-7)^2+15)du`

Apply the substitution `8u-7=sqrt15tantheta`:

`L=1/2(sqrt265-1)-int((7-64/(sqrt15tantheta+7)))/(sqrt15sectheta)(sqrt15/8sec^2thetad theta)`

Simplify and distribute:

`L=1/2(sqrt265-1)-7/8intsecthetad theta+8intsectheta/(sqrt15tantheta+7)d theta`

Apply the appropriate double-angle Trigonometric identities:

`L=1/2(sqrt265-1)-7/8[ln|sqrt15sectheta+sqrt15tantheta|]+8intsec^2(theta/2)/(2sqrt15tan(theta/2)+7(1-tan^2(theta/2)))d theta`

Rearrange:

`L=1/2(sqrt265-1)-7/8[ln|sqrt((8u-7)^2+15)+8u-7|]_ 1^9-8intsec^2(theta/2)/(7tan^2(theta/2)-2sqrt15tan(theta/2)-7)d theta`

Complete the square in the denominator:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-56intsec^2(theta/2)/((7tan(theta/2)-sqrt15)^2-64)d theta`

Apply the difference of squares:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-56intsec^2(theta/2)/((7tan(theta/2)-sqrt15-8)(7tan(theta/2)-sqrt15+8))d theta`

Apply partial fraction decomposition:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-7/2int(1/(7tan(theta/2)-sqrt15-8)-1/(7tan(theta/2)-sqrt15+8))sec^2(theta/2)d theta`

Integrate term by term:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-[ln|7tan(theta/2)-sqrt15-8|-ln|7tan(theta/2)-sqrt15+8|]`

Apply the appropriate half-angle Trigonometric identity:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-[ln|(7sqrt15tantheta-(sqrt15+8)(sqrt15+sqrt15sectheta))/(7sqrt15tantheta-(sqrt15-8)(sqrt15+sqrt15sectheta))|]`

Reverse the last substitution:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-[ln|(7(8u-7)-(sqrt15+8)(sqrt15+sqrt((8u-7)^2+15)))/(7(8u-7)-(sqrt15-8)(sqrt15+sqrt((8u-7)^2+15)))|]_1^9`

Insert the limits of integration:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-ln|(455-(sqrt15+8)(sqrt15+4sqrt265))/(455-(sqrt15-8)(sqrt15+4sqrt265))*(7-(sqrt15-8)(sqrt15+4))/(7-(sqrt15+8)(sqrt15+4))|`

Simplify:

`L=1/2(sqrt265-1)-7/8ln((4sqrt265+65)/5)-ln((4sqrt265-55)/45)`