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

[tex]f) \int(x + 1)lnx \: dx[/tex]

[tex]f = lnx = > f' = \frac{1}{x} [/tex]

[tex]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[/tex]

[tex] \int f \times g' \: dx = f \times g - \int f' \times g \: dx[/tex]

[tex] \int(x + 1)lnx \: dx = lnx \times \frac{ {x}^{2} + 2x}{2} - \int \frac{1}{x} \times \frac{ {x}^{2} + 2x}{2} \: dx[/tex]

[tex] = lnx \times \frac{ {x}^{2} + 2x }{2} - \frac{1}{2} \int \frac{ {x}^{2} + 2x }{x} \: dx[/tex]

[tex] = lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \int \frac{x(x + 2)}{x} \: dx[/tex]

[tex] = lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \int(x + 2) \: dx[/tex]

[tex] = lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \int x \: dx - \frac{1}{2} \int 2 \: dx[/tex]

[tex] = lnx \times \frac{ {x}^{2} + 2x}{2} - \frac{1}{2} \times \frac{ {x}^{2} }{2} - \frac{1}{2} \times2 \int1 \: dx[/tex]

[tex] = lnx \times \frac{ {x}^{2} + 2x }{2} - \frac{ {x}^{2} }{4} - x + C[/tex]

[tex]i) \int(3 {x}^{2} + 2x) {e}^{x} \: dx[/tex]

[tex]f = 3 {x}^{2} + 2x [/tex]

[tex]f' = (3 {x}^{2} + 2x)' [/tex]

[tex]f' = (3 {x}^{2} ) ' + (2x)'[/tex]

[tex]f' = 3 \times ( {x}^{2} )' + 2 \times (x)'[/tex]

[tex]f' = 3 \times 2x + 2 \times 1[/tex]

[tex]f' = 6x + 2[/tex]

[tex]g' = {e}^{x} = > g = \int {e}^{x} \: dx = {e}^{x} +C[/tex]

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

[tex]f = 6x + 2[/tex]

[tex]f' = (6x + 2)' [/tex]

[tex]f' = (6x)' + 2'[/tex]

[tex]f' = 6 \times (x)' + 0[/tex]

[tex]f' = 6 \times 1[/tex]

[tex]f' = 6[/tex]

[tex]g' = {e}^{x} = > g = \int {e}^{x} \: dx = {e}^{x} + C[/tex]

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

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

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

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

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

[tex] = {e}^{x} (3 {x}^{2} - 4x + 4) + C[/tex]