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education/math/MATH1220 (calc II)/Integration by Parts.md

@@ -5,6 +5,7 @@ $$ \int udv = uv - \int vdu $$
 $$ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) $$
 1. Integrating both sides, we get:
 $$\int \frac{d}{dx} (f(x)g(x))dx = \int [f'(x)g(x) + f(x)]$$
-
-2. Therefore:
-$$$$
+2. Through the distributive property of integrals,
+$$ = \int f'(x)g(x)dx + \int f(x)g'(x)dx $$
+3. Therefore:
+$$f(x)g(x) = \intf'(x)g(x)dx $$