Ka mau tanya. tolong bantuin kerjain pusing aku udah

Ka mau tanya. tolong bantuin kerjain pusing aku udah

Penjelasan dengan langkah-langkah:

$int(3x - 5)(x + 3)dx \ =int( 3 {x}^{2} + 9x - 5x - 15)dx \ = frac{3}{3} {x}^{3} + frac{9}{2} {x}^{2} - frac{5}{2} {x}^{2} - 15x + c \ = {x}^{3} + 2 {x}^{2} - 15x + c$

$int( frac{6 {x}^{4} - 4 {x}^{3} + 2 {x}^{2} - 1}{ {x}^{2} } )dx \ =int (6 {x}^{2} - 4x + 2 - frac{1}{ {x}^{2} } )dx \ = frac{6}{3} {x}^{3} - frac{4}{2} {x}^{2} + 2x - frac{1}{ - 2 + 1} {x}^{ - 2 + 1} + c \ = 2 {x}^{3} - 2 {x}^{2} + 2x + frac{1}{x} + c$

$y = ( frac{dy}{dx} )dx \ = int (3 {x}^{2} + 4x - 2)dx \ = frac{3}{3} {x}^{3} + frac{4}{2} {x}^{2} - 2x + c \ = {x}^{3} + 2 {x}^{2} - 2x + c$

melalui (-1,5) substitusi

$5 = ( { - 1)}^{3} + 2( { - 1)}^{2} - 2( - 1) + c \ 5 = - 1 + 2 + 2 + c \ 5 = 3 + c \ c = 5 - 3 = 2$

maka persamaan

$y = {x}^{3} + 2 {x}^{2} - 2x + 2$

$int limits_{0}^{2}(x(3x + 5))dx \ =intlimits_{0}^{2}( 3 {x}^{2} + 5x)dx \ = frac{3}{3} {x}^{3} + frac{5}{2} {x}^{2} \ = {x}^{3} + frac{5}{2} {x}^{2} \ = {2}^{3} + frac{5}{2} ( {2)}^{2} - 0 \ = 8 + 10 = 18$

$intlimits_{-1}^{2}(6 {x}^{2} + 8x)dx \ = frac{6}{3} {x}^{3} + frac{8}{2} {x}^{2} \ = 2 {x}^{3} + 4 {x}^{2} \ = 2( {2)}^{3} + 4( {2)}^{2} - 2( { - 1)}^{3} - 4( { - 1)}^{2} \ = 16 + 16 + 2 - 4 \ = 30$

$intlimits_{0}^{3}( {x}^{2} + px + 2)dx =frac{3}{2} \ frac{1}{3} {x}^{3} + frac{p}{2} {x}^{2} + 2x = frac{3}{2} \ frac{1}{3} ( {3)}^{3} + frac{p}{2} ( {3)}^{2} + 2(3) = frac{3}{2} \ 9 + frac{9}{2} p + 6 = frac{3}{2} \ frac{9}{2} p + 15 = frac{3}{2} \ 9p + 30 = 3 \ 9p = 3 - 30 \ 9p = - 27 \ p = frac{ - 27}{9} = - 3$

$u = 2x + 9 \ du = 2dx \ dx = frac{du}{2}$

$int{u}^{ frac{1}{2} } ( frac{du}{2} ) \ = frac{1}{2} ( frac{1}{ frac{1}{2} + frac{2}{2} } {u}^{ frac{1}{2} + frac{2}{2} } ) + c \ = frac{1}{2} ( frac{1}{ frac{3}{2} } {u}^{ frac{3}{2} } ) + c \ = frac{1}{3} (2x + 9) sqrt{2x + 9} + c$

$u = 2x + 3 \ du = 2dx \ dx = frac{du}{2}$

$int 12 {u}^{ - 2} ( frac{du}{2} ) \ = 6( frac{1}{ - 2 + 1} {u}^{ - 2 + 1} ) + c \ = - 6 {u}^{ - 1} + c \ = frac{ - 6}{2x + 3} + c$

$u = 4x - 5 \ du = 4dx \ dx = frac{du}{4}$

$int 2 {u}^{ - frac{1}{2} } ( frac{du}{4} ) \ = frac{1}{2} ( frac{1}{ - frac{1}{2} + frac{2}{2} } {u}^{ - frac{1}{2} + frac{2}{2} } ) + c \ = frac{1}{2} ( frac{1}{ frac{1}{2} } ) {u}^{ frac{1}{2} } + c \ = sqrt{4x - 5} + c$

10)

$u = {x}^{3} + 2 \ du = 3 {x}^{2} dx \ dx = frac{du}{3 {x}^{2} }$

$int 2 {x}^{2} ( {u}^{5} ) frac{du}{3 {x}^{2} } \ = frac{2}{3} ( frac{1}{5 + 1} ) {u}^{5 + 1} + c \ frac{1}{9} {u}^{6} + c \ = frac{1}{9} ( {x}^{3} + 2) ^{6} + c$