Integration is the reverse process of differentiation. Antiderivatives allow you to recover an original function from its derivative, and are key to calculating area under curves, motion from acceleration, and accumulated change.
F(x)
such that F'(x) = f(x)
\int f(x) dx = F(x) + C
, where C
is the constant of integration\int x^n dx = \frac{x^{n+1}}{n+1} + C
(for n ≠ -1
)\int \frac{1}{x} dx = \ln|x| + C
\int e^x dx = e^x + C
; \int a^x dx = \frac{a^x}{\ln a} + C
\int \sin x dx = -\cos x + C
; \int \cos x dx = \sin x + C
u = g(x)
so du = g'(x) dx
.\frac{d}{dx}[\int f(x) dx] = f(x)
, you’re correct!
f(x) = 6x^2
\int 6x^2 dx = 6 \cdot \int x^2 dx = 6 \cdot \frac{x^3}{3} + C = 2x^3 + C
Final Answer: 2x^3 + C
\int (2x)(x^2 + 1)^5 dx
using substitution
Let u = x^2 + 1
→ du = 2x dx
So the integral becomes \int u^5 du = \frac{u^6}{6} + C
Substitute back: \frac{(x^2 + 1)^6}{6} + C
Final Answer: \frac{(x^2 + 1)^6}{6} + C