Question

Vă rog ajutațimă la cele doua subpuncte. Rezolvare completa dacă se poate

Answer 1

$f) \int(x + 1)lnx \: dx$

$f = lnx = > f' = \frac{1}{x}$

$g' = x + 1 = > g = \int(x + 1) \: dx = \int x \: dx + \int 1 \: dx = \frac{ {x}^{2} }{2} + x = \frac{ {x}^{2} + 2x }{2} + C$

$\int f \times g' \: dx = f \times g - \int f' \times g \: dx$

$\int(x + 1)lnx \: dx = lnx \times \frac{ {x}^{2} + 2x}{2} - \int \frac{1}{x} \times \frac{ {x}^{2} + 2x}{2} \: dx$

$= lnx \times \frac{ {x}^{2} + 2x }{2} - \frac{1}{2} \int \frac{ {x}^{2} + 2x }{x} \: dx$

$= lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \int \frac{x(x + 2)}{x} \: dx$

$= lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \int(x + 2) \: dx$

$= lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \int x \: dx - \frac{1}{2} \int 2 \: dx$

$= lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \times \frac{ {x}^{2} }{2} - \frac{1}{2} \times2 \int1 \: dx$

$= lnx \times \frac{ {x}^{2} + 2x }{2} - \frac{ {x}^{2} }{4} - x + C$

$i) \int(3 {x}^{2} + 2x) {e}^{x} \: dx$

$f = 3 {x}^{2} + 2x$

$f' = (3 {x}^{2} + 2x)'$

$f' = (3 {x}^{2} ) ' + (2x)'$

$f' = 3 \times ( {x}^{2} )' + 2 \times (x)'$

$f' = 3 \times 2x + 2 \times 1$

$f' = 6x + 2$

$g' = {e}^{x} = > g = \int {e}^{x} \: dx = {e}^{x} +C$

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

$f = 6x + 2$

$f' = (6x + 2)'$

$f' = (6x)' + 2'$

$f' = 6 \times (x)' + 0$

$f' = 6 \times 1$

$f' = 6$

$g' = {e}^{x} = > g = \int {e}^{x} \: dx = {e}^{x} + C$

$\int(3 {x}^{2} + 2x) {e}^{x} = (3 {x}^{2} + 2x) {e}^{x} - [(6x + 2) {e}^{x} - \int6 {e}^{x} \: dx ]$

$= (3 {x}^{2} + 2x) {e}^{x} - [(6x + 2) {e}^{x} - 6\int {e}^{x} \: dx ]$

$= (3 {x}^{2} + 2x) {e}^{x} - [(6x + 2) {e}^{x} - 6 {e}^{x} ]$

$= (3 {x}^{2} + 2x) {e}^{x} - (6x + 2) {e}^{x} + 6 {e}^{x}$

$= {e}^{x} (3 {x}^{2} + 2x -6 x - 2 + 6)$

$= {e}^{x} (3 {x}^{2} - 4x + 4) + C$