# What is the indefinite integral of y=3x^2+2x-1 ?

*print*Print*list*Cite

### 3 Answers

We have to find the integral of y=3x^2+2x-1.

Now Int [y] = Int [ 3x^2+2x-1]

=> Int [ 3x^2] + Int [ 2x] - Int [1]

=> 3*x^3 / 3 + 2* x^2 / 2 - x + C

=> x^3 + x^2 - x + C

**Therefore the required integral is x^3 + x^2 - x + C**

To find the indefinite integral of y=3x^2+2x-1.

Let f(x) = 3x^2+2x-1.

Then Int f(x) = Int (3x^2+2x-1)dx.

Int f(x) = Int3x^2 dx + Int 2x dx - Int 1*dx

Intf(x) = 3*(1/2+1) x^(2+1) + 2* (1/(1+1)x^(1+1) -x +C.

Int f(x) dx = x^3 + x^2 - x +C.

We'll write y = f(x) and we'll compute the indefinite integral of f(x):

Int f(x)dx = Int (3x^2 + 2x -1)dx

We'll apply the property of integral to be additive and we'll get:

Int (3x^2 + 2x -1)dx = Int 3x^2dx + Int 2xdx - Int dx

We'll re-write the sum of integrals:

Int (3x^2 + 2x -1)dx = 3Int x^2dx + 2Int xdx - Int dx

Int (3x^2 + 2x -1)dx = 3*x^3/3 + 2*x^2/2 - x + C

We'll simplify and we'll get the indefinite integral of y=3x^2+2x-1:

**Int (3x^2 + 2x -1)dx = x^3 + x^2 - x + C**