giúp mình câu này nha các tềnh iu tính nguyên hàm các hàm số sau: a/ 2x+3 b/ 4x+1 c/ 3x^2+2x+6 d/ e^x+1/(x+3)

a) $∫(2x+3)dx=x^2+3x+C$

b) $∫(4x+1)dx=2x^2+x+C$

c) $∫(3x^2+2x+6)dx=x^3+x^2+6x+C$

d) $∫(e^x+\frac{1}{x+3})dx=e^x+ln|x+3|+C$.

a,

$\displaystyle\int (2x+3)dx$

$=\displaystyle\int 2xdx+\displaystyle\int 3dx$

$=x^2+3x+C$

b,

$\displaystyle\int (4x+1)dx$

$=\displaystyle\int 4xdx+\displaystyle dx$

$=2x^2+x+C$

c,

$\displaystyle\int (3x^2+2x+6)dx$

$=\displaystyle\int 3x^2dx+\displaystyle\int 2xdx+\displaystyle\int 6dx$

$=x^3+x^2+6x+C$

d,

$\displaystyle\int \left(e^x+\dfrac{1}{x+3}\right)dx$

$=\displaystyle\int e^xdx+\displaystyle\int \dfrac{dx}{x+3}$

$=e^x+\ln|x+3|+C$