Issolvi ∫ left(4x-2right)^3 wrt x

$$\int{ { \left(4x-2 \right) }^{ 3 } }d x$$

Evalwa

$$\left(2x-1\right)^{4}+С$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} biex tespandi \left(4x-2\right)^{3}.

$$\int 64x^{3}-96x^{2}+48x-8\mathrm{d}x$$

Tintegra s-somma terminu b'terminu.

$$\int 64x^{3}\mathrm{d}x+\int -96x^{2}\mathrm{d}x+\int 48x\mathrm{d}x+\int -8\mathrm{d}x$$

Iffattura ‘l barra l-kostanti f’kull wieħed minn dawn it-termini.

$$64\int x^{3}\mathrm{d}x-96\int x^{2}\mathrm{d}x+48\int x\mathrm{d}x+\int -8\mathrm{d}x$$

Minn \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} għal k\neq -1, issostitwixxi \int x^{3}\mathrm{d}x ma' \frac{x^{4}}{4}. Immultiplika 64 b'\frac{x^{4}}{4}.

$$16x^{4}-96\int x^{2}\mathrm{d}x+48\int x\mathrm{d}x+\int -8\mathrm{d}x$$

Minn \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} għal k\neq -1, issostitwixxi \int x^{2}\mathrm{d}x ma' \frac{x^{3}}{3}. Immultiplika -96 b'\frac{x^{3}}{3}.

$$16x^{4}-32x^{3}+48\int x\mathrm{d}x+\int -8\mathrm{d}x$$

Minn \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} għal k\neq -1, issostitwixxi \int x\mathrm{d}x ma' \frac{x^{2}}{2}. Immultiplika 48 b'\frac{x^{2}}{2}.

$$16x^{4}-32x^{3}+24x^{2}+\int -8\mathrm{d}x$$

Sib l-integru ta' -8 billi tuża t-tabella tar-regola tal-integrali komuni \int a\mathrm{d}x=ax.

$$16x^{4}-32x^{3}+24x^{2}-8x$$

Jekk F\left(x\right) huwa antiderivattiv ta' f\left(x\right), allura s-sett tal-antiderivati kollha ta' f\left(x\right) jingħata minn F\left(x\right)+C. Għalhekk, żid il-kostanti ta' integrazzjoni C\in \mathrm{R} mar-riżultat.

$$16x^{4}-32x^{3}+24x^{2}-8x+С$$

Iddifferenzja w.r.t. x

$$8\left(2x-1\right)^{3}$$