Tolong bantu jawab

Tolong bantu jawab

$int_{ - 1 }^{3} {(2x + 3)}^{ - 2} dx \$

misal

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

rubah batas integral, substitusi ke u = 2x+3

$- 1 to : 2( - 1) + 3 = 1 \ 3 to : 2(3) + 3 = 9$

bentuk integralnya menjadi

$int_{ - 1 }^{3} {(2x + 3)}^{ - 2} dx \ = int_{ 1}^{9} {(u)}^{ - 2} frac{du}{2} \ = frac{1}{2} ( frac{1}{ - 2 + 1} {u}^{ - 2 + 1} )| _{ 1 }^{9} \ = frac{1}{2} ( - {u}^{ - 1} ) | _{ 1 }^{9} \ = frac{1}{2} ( - frac{1}{u} )| _{ 1 }^{9} \ = - frac{1}{2} ( frac{1}{9} - 1 ) \ = - frac{1}{2} ( - frac{8}{9} ) \ = frac{4}{9}$