diff --git a/education/math/MATH1210 (calc 1)/Integrals.md b/education/math/MATH1210 (calc 1)/Integrals.md
index 3b866ef..40e50d2 100644
--- a/education/math/MATH1210 (calc 1)/Integrals.md	
+++ b/education/math/MATH1210 (calc 1)/Integrals.md	
@@ -162,6 +162,10 @@ $$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
 5. $\frac{13}{12} x \Big| _{-1}^8$
 
 > Find the distance from the point ${\frac{1}{2}, \frac{49}{48}}$ to the point $(5, \frac{314}{15})$ along the curve $y = \dfrac{x^4 - 3}{6x}$
-1. $y' = \dfrac{4x^3(6x) - (x^4 + 3)6}{36x^2}$
-2. $= \dfrac{24x^4 -6x^4 - 18}{36x^2}$
-3. 
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+1. $y' = \dfrac{4x^3(6x) - (x^4 + 3)6}{36x^2}$: Find the derivative of the curve using the quotient rule
+2. $= \dfrac{18x^4 - 18}{36x^2}$: Simplify
+3. $= \dfrac{18(x^4 - 1)}{18(2x^2)}$: Factor out $18$
+4. $= \dfrac{x^4 - 1}{2x^2}$: Factor out $18$ again
+5. $L = \int_{1/2}^5 \sqrt{1 + (\dfrac{4x-1}{2x^2})^2}dx$ : Use the length formula
+6. $= \int_{1/2}^5 \sqrt{1 + \dfrac{x^8 - 2x^4 + 1}{x^4}} dx$: Apply the $^2$ 
+7. $= \int_{1/2}^5 \sqrt{\dfrac{4x^4 + x^8 -2x^4 + 1}{4x^4}}$: Set $1 = \dfrac{4x^4}{4x^4}$ and add
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