hasil dari diatas................................

Materi : Kalkulus
Bahasan : Integral Aljabar Tak Tentu
Jawaban : E.

Langkah - langkah :
[tex] \int\limits {(4x^3 + \frac{1}{2}x^2 + 3x )} \, dx \\ \\ = 4.\frac{x^{3 + 1}}{3 + 1} + \frac{1}{2}. \frac{x^{2 + 1}}{2 + 1} + 3. \frac{x^{1 + 1}}{1 + 1} + C \\ \\ = 4. \frac{x^4}{4} + \frac{1}{2} . \frac{x^3}{3} + 3 . \frac{x^2}{2} + C \\ \\ = x^4 + \frac{x^3}{6} + \frac{3x^2}{2} + C \\ \\ = x^4 + \frac{1}{6}x^3 + \frac{3}{2}x^2 + C [/tex]