Riješi ∫ x^2(x+1)^3 wrt x

$$\int x ^ { 2 } ( x + 1 ) ^ { 3 } d x$$

Izračunaj

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+\frac{3x^{4}}{4}+\frac{x^{3}}{3}+С$$

Koraci rješenja

Upotrijebite binomni teorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} da biste proširili \left(x+1\right)^{3}.

$$\int x^{2}\left(x^{3}+3x^{2}+3x+1\right)\mathrm{d}x$$

Koristite svojstvo distributivnosti da biste pomnožili x^{2} s x^{3}+3x^{2}+3x+1.

$$\int x^{5}+3x^{4}+3x^{3}+x^{2}\mathrm{d}x$$

Integrirajte zbroj termina po terminu.

$$\int x^{5}\mathrm{d}x+\int 3x^{4}\mathrm{d}x+\int 3x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Izbacite konstantu u svakom od izraza.

$$\int x^{5}\mathrm{d}x+3\int x^{4}\mathrm{d}x+3\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Od \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} za k\neq -1, zamijenite \int x^{5}\mathrm{d}x s \frac{x^{6}}{6}.

$$\frac{x^{6}}{6}+3\int x^{4}\mathrm{d}x+3\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Od \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} za k\neq -1, zamijenite \int x^{4}\mathrm{d}x s \frac{x^{5}}{5}. Pomnožite 3 i \frac{x^{5}}{5}.

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+3\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Od \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} za k\neq -1, zamijenite \int x^{3}\mathrm{d}x s \frac{x^{4}}{4}. Pomnožite 3 i \frac{x^{4}}{4}.

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+\frac{3x^{4}}{4}+\int x^{2}\mathrm{d}x$$

Od \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} za k\neq -1, zamijenite \int x^{2}\mathrm{d}x s \frac{x^{3}}{3}.

$$\frac{x^{6}}{6}+\frac{3x^{5}}{5}+\frac{3x^{4}}{4}+\frac{x^{3}}{3}$$

Pojednostavnite.

$$\frac{x^{3}}{3}+\frac{3x^{4}}{4}+\frac{3x^{5}}{5}+\frac{x^{6}}{6}$$

Ako je F\left(x\right) antiderivat f\left(x\right), skup svih antiderivata f\left(x\right) je dan u F\left(x\right)+C. Stoga dodajte konstantu integracije C\in \mathrm{R} rezultatu.

$$\frac{x^{3}}{3}+\frac{3x^{4}}{4}+\frac{3x^{5}}{5}+\frac{x^{6}}{6}+С$$

Diferenciraj u odnosu na x

$$x^{2}\left(x+1\right)^{3}$$