Datrys ∫ x^2(x+1)^3 wrt x

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

Enrhifo

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

Camau Ateb

Defnyddio'r theorem binomaidd \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} i ehangu'r \left(x+1\right)^{3}.

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

Defnyddio’r briodwedd ddosbarthu i luosi x^{2} â x^{3}+3x^{2}+3x+1.

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

Integreiddio'r swm fesul term.

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

Ffactoreiddio allan y cysonyn ym mhob un o'r termau.

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

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{5}\mathrm{d}x gyda \frac{x^{6}}{6}.

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

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{4}\mathrm{d}x gyda \frac{x^{5}}{5}. Lluoswch 3 â \frac{x^{5}}{5}.

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

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{3}\mathrm{d}x gyda \frac{x^{4}}{4}. Lluoswch 3 â \frac{x^{4}}{4}.

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

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{2}\mathrm{d}x gyda \frac{x^{3}}{3}.

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

Symleiddio.

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

Os yw F\left(x\right) yn integryn amhendant o f\left(x\right), yna bydd F\left(x\right)+C yn rhoi’r set o holl integrynnau amhendant f\left(x\right). Felly, ychwanegwch gysonyn yr integryn C\in \mathrm{R} at y canlyniad.

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

Gwahaniaethu w.r.t. x

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