Integrate the following w.r.t. x.
1. `x^3`
2. `x-1/x`
3. `e^(2x)+(1)/(x^2)`

Options

A. `(x^4)/(4)+c` , `x^2/2-x+c` ,`(e^(2x))/(2)-1/x+c`
B. `(x^4)+c` , `x^2/2-1nx+c` ,`(e^(2x))/(2)-1/x+c`
C. `(x^4)/(4)+c` , `x^2/2-1nx+c` ,`(e^(2x))/(2)-1/x+c`
D. `(x^4)/(4)+c` , `x^2/2-1nx+c` ,`(e^(2x))/(2)-1+c`

Answer

C

Explanation

1. `y=x^3`
Integrating both sides w.r.t. x, we get
`I=intydx=intx^3dx=(x^4)/(4)+c[:'intx^ndx=(x^(n+1))/(n+1)+c]`
2. `y=x-1/x`
Integrating both sides w.r.t. x, we get
`I=intydx=int(x-1/x)dx[:'int1/xdx=1nx+c]`
`=x^2/2-1nx+c`
3. `y=e^(2x)+(1)/(x^2)`
Integrating both sides w.r.t. x, we get
`I=intydx=int(e^(2x)+(1)/(x^2))dx`
`=(e^(2x))/(2)+(x^(-2+1))/(-2+1)+c[:'inte^(ax)=(e^(ax))/(a)+c]`
`=(e^(2x))/(2)-1/x+c`
4. `y=(1)/(2x+3)`
Integrating both sides w.r.t. x, we get
`I=intydx=int(dx)/(2x+3)+(1n|2x+3|)/(2)+c`
`y=cos(4x+3)`
Integrating both sides w.r.t. x, we get
`I=intydx=intcos(4x+3)dx=(sin(4x+3))/(4)+c`
6. `y=cos^2x`
Integrating both sides w.r.t. x, we get
`I=intydx`
`=intcos^2xdx`
`=int((1+cos2x)/(2))dx`
`=int1/2dx+int(cos2x)/(2)dx`
`=x/2+(sin2x)/(2xx2)+c=1/2(x+(sin2x)/(2))+c`