Question

alguien que me ayude plis ​

Answer 1

$\int3 {x}^{2} dx \\ = 3 \frac{ {x}^{3} }{3} \\ = {x}^{3} + c$

$\int \frac{dy}{ \sqrt{y} } \\ \\ = \int {y}^{ \frac{ - 1}{2} } dy \\ \\ = \frac{ {y}^{ \frac{1}{2} } }{ \frac{1}{2} } \\ \\ = 2 \sqrt{y} + c$

$\int(2x + 4 {x}^{2} - 6 {x}^{3} )dx \\ \\ = \int2x \: dx + \int4 {x}^{2} \: dx - \int6 {x}^{3} \: dx \\ \\ = (2 \times \frac{ {x}^{2} }{2}) + (4 \times \frac{ {x}^{3} }{3} ) - (6 \times \frac{ {x}^{4} }{4} ) \\ \\ = {x}^{2} + \frac{4}{3} {x}^{3} - \frac{3}{2} {x}^{4} + c$

$\int {( {x}^{3} - 5) }^{3} 6 {x}^{2} \: dx \\ \\ sea \: u = {x}^{3} - 5 \\ du = 3 {x}^{2} dx \\ \\ \int {u}^{3} 2du \\ \\ = (2 \times \frac{ {u}^{4} }{4} ) \\ \\ = \frac{ {( {x}^{3} - 5)}^{4} }{2} + c$

$\int \sqrt{4x + 7} \: dx \\ \\ u = 4x + 7 \\ du = 4dx \\ \frac{du}{4} = dx \\ \\ \int \sqrt{u} \frac{du}{4} \\ \\ = \frac{1}{4} \int {u}^{ \frac{1}{2} } du \\ \\ = \frac{1}{4} ( \frac{ {u}^{ \frac{3}{2} } }{ \frac{3}{2} } ) \\ \\ = \frac{ {u}^{ \frac{3}{2} } }{6} \\ \\ = \frac{ {(4x + 7)}^{ \frac{3}{2} } }{6} + c$

Las otras son fáciles solo que es difícil escribirlas por cada, espero le sirvan