Question

Question 1 Integrate each of the following:
(a) x² - 2x + 1/x²
(b) x(x + 3)²​

Answer 1

Explanation:

I've skipped some steps. Hope you get this.

Answer 2

(a)
`\displaystyle{\int \left(x^2-2x+(1)/(x^2)\right) \, dx} = (x^3)/(3) - x^2 - (1)/(x) + \text{C}}`

(b)
`\displaystyle{\int x\left(x+3\right)^2 \, dx = (x^4)/(4) + 2x^3 + (9x^2)/(2) + \text{C}}`

Explanation:

Part A

Consider the indefinite integral with respect to x:

`\displaystyle{\int \left(x^2-2x+(1)/(x^2)\right) \, dx}`

We can separately integrate each terms as they are in addition and subtraction:

`\displaystyle{\int x^2 \, dx - \int 2x \, dx + \int (1)/(x^2) \, dx}`

For the last term, we can rewrite in negative exponent as
`x^(-2)` so we can apply power rules to all these terms (making it easier):

`\displaystyle{\int x^2 \, dx - \int 2x \, dx + \int x^(-2) \, dx}`

The middle term -- we can separate the constant out of integral:

`\displaystyle{\int x^2 \, dx - 2\int x \, dx + \int x^(-2) \, dx}`

Apply the power integration rule (for n ≠ 1):

`\displaystyle{\int x^n \, dx = (x^(n+1))/(n+1) + \text{C} \ \ \left(\text{where C is a constant}\right)}`

But we can add C later after we integrate all these terms, therefore:

`\displaystyle{(x^(2+1))/(2+1) - 2\cdot (x^(1+1))/(1+1) + (x^(-2+1))/(-2+1)}\\\\\displaystyle{(x^(3))/(3) - 2\cdot (x^(2))/(2) + (x^(-1))/(-1)}\\\\\displaystyle{(x^3)/(3) - x^2 - x^(-1)}`

Rewrite the negative exponent back to fraction form:

`\displaystyle{(x^3)/(3) - x^2 - (1)/(x)}`

After integrating all terms, we add + C every time. Hence:

`\displaystyle{\int \left(x^2-2x+(1)/(x^2)\right) \, dx} = (x^3)/(3) - x^2 - (1)/(x) + \text{C}}`

Part B

Consider the indefinite integration with respect to x of:

`\displaystyle{\int x\left(x+3\right)^2 \, dx}`

First, expand (x + 3)² to x² + 6x + 9:

`\displaystyle{\int x\left(x^2+6x+9\right) \, dx}`

Then disribute the x inside the brackets:

`\displaystyle{\int \left(x^3+6x^2+9x\right) \, dx}`

Integrate separately:

`\displaystyle{\int x^3 \, dx + \int 6x^2 \, dx + \int 9x \, dx}`

Separate the constants:

`\displaystyle{\int x^3 \, dx + 6\int x^2 \, dx + 9\int x \, dx}`

For the power integration formula, recall above. Thus:

`\displaystyle{(x^(3+1))/(3+1) + 6 \cdot (x^(2+1))/(2+1) + 9 \cdot (x^(1+1))/(1+1)}\\\\\displaystyle{(x^(4))/(4) + 6 \cdot (x^(3))/(3) + 9 \cdot (x^(2))/(2)}\\\\\displaystyle{(x^4)/(4) + 2x^3 + (9x^2)/(2)}`

Add + C after finishing the integration process, hence:

`\displaystyle{\int x\left(x+3\right)^2 \, dx = (x^4)/(4) + 2x^3 + (9x^2)/(2) + \text{C}}`

Let me know if you have any questions!