Datrys ∫ (4x^7+4x+4)(28x^6+4) wrt x

$$\int ( 4 x ^ { 7 } + 4 x + 4 ) ( 28 x ^ { 6 } + 4 ) d x$$

Enrhifo

$$8x^{14}+16x^{8}+16x^{7}+8x^{2}+16x+С$$

Camau Ateb

Defnyddio’r briodwedd ddosbarthu i luosi 4x^{7}+4x+4 â 28x^{6}+4 a chyfuno termau tebyg.

$$\int 112x^{13}+128x^{7}+16x+112x^{6}+16\mathrm{d}x$$

Integreiddio'r swm fesul term.

$$\int 112x^{13}\mathrm{d}x+\int 128x^{7}\mathrm{d}x+\int 16x\mathrm{d}x+\int 112x^{6}\mathrm{d}x+\int 16\mathrm{d}x$$

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

$$112\int x^{13}\mathrm{d}x+128\int x^{7}\mathrm{d}x+16\int x\mathrm{d}x+112\int x^{6}\mathrm{d}x+\int 16\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{13}\mathrm{d}x gyda \frac{x^{14}}{14}. Lluoswch 112 â \frac{x^{14}}{14}.

$$8x^{14}+128\int x^{7}\mathrm{d}x+16\int x\mathrm{d}x+112\int x^{6}\mathrm{d}x+\int 16\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{7}\mathrm{d}x gyda \frac{x^{8}}{8}. Lluoswch 128 â \frac{x^{8}}{8}.

$$8x^{14}+16x^{8}+16\int x\mathrm{d}x+112\int x^{6}\mathrm{d}x+\int 16\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x\mathrm{d}x gyda \frac{x^{2}}{2}. Lluoswch 16 â \frac{x^{2}}{2}.

$$8x^{14}+16x^{8}+8x^{2}+112\int x^{6}\mathrm{d}x+\int 16\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{6}\mathrm{d}x gyda \frac{x^{7}}{7}. Lluoswch 112 â \frac{x^{7}}{7}.

$$8x^{14}+16x^{8}+8x^{2}+16x^{7}+\int 16\mathrm{d}x$$

Canfod integryn 16 gan ddefnyddio'r rheol tabl o integrynnau cyffredin \int a\mathrm{d}x=ax.

$$8x^{14}+16x^{8}+8x^{2}+16x^{7}+16x$$

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.

$$8x^{14}+16x^{8}+16x^{7}+8x^{2}+16x+С$$

Gwahaniaethu w.r.t. x

$$16\left(7x^{6}+1\right)\left(x^{7}+x+1\right)$$