Question

Sa se calculeze integrala nedifinita:
$\int\limits {(x^2-2x)^3} \, dx$

Answer 1

$\int {( {x}^{2} - 2x)}^{3} \: dx$

$= \int[ { ({x}^{2} )}^{3} - 3 {( {x}^{2}) }^{2} \times 2x + 3 {x}^{2} \times ( {2x})^{2} -( {2x})^{3} ] \: dx$

$= \int ({x}^{6} - 3 {x}^{4} \times 2x + 3 {x}^{2} \times 4 {x}^{2} - 8 {x}^{3} ) \: dx$

$= \int( {x}^{6} - 6 {x}^{5} + 12 {x}^{4} - 8 {x}^{3} ) \: dx$

$= \int {x}^{6} \: dx - \int6 {x}^{5} \: dx + \int12 {x}^{4} \: dx - \int8 {x}^{3} \: dx$

$= \int {x}^{6} \: dx - 6 \int {x}^{5} \: dx + 12 \int {x}^{4} \: dx - 8 \int {x}^{3} \: dx$

$= \frac{ {x}^{6 + 1} }{6 + 1} - 6 \times \frac{ {x}^{5 + 1} }{5 + 1} + 12 \times \frac{ {x}^{4 + 1} }{4 + 1} - 8 \times \frac{ {x}^{3 + 1} }{3 + 1}$

$= \frac{ {x}^{7} }{7} - 6 \times \frac{ {x}^{6} }{6} + 12 \times \frac{ {x}^{5} }{5} - 8 \times \frac{ {x}^{4} }{4}$

$= \frac{ {x}^{7} }{7} - {x}^{6} + \frac{12 {x}^{5} }{5} - 2 {x}^{4}$

$= \frac{5 {x}^{7} - 35 {x}^{6} + 84 {x}^{5} - 70 {x}^{4} }{35}$

$= \frac{ {x}^{4} (5 {x}^{3} - 35 {x}^{2} + 84 x - 70) }{35} + C$