Question

Evaluate the attached integral. The answer I got was
`2x^2 -x+5ln\left|x-2\right|+(1)/(3)\left(x-2\right)^3-6+C` , I'm not entirely sure if my method was correct though.

Answer 1

`=2x^2-x+5\ln \left|x-2\right|+(1)/(3)\left(x-2\right)^3-6+C`

Explanation:

`\int (x^3-2x^2+3x-1)/(x-2)dx`

`(x^3-2x^2+3x-1)/(x-2)\\`

`=(x^3)/(x-2)-(2x^2)/(x-2)+(3x)/(x-2)-(1)/(x-2)`

`=\int (x^3)/(x-2)dx-\int (2x^2)/(x-2)dx+\int (3x)/(x-2)dx-\int (1)/(x-2)dx`

`\int (x^3)/(x-2)dx`

`=\int (\left(u+2\right)^3)/(u)du; u=x-2`

`(\left(u+2\right)^3)/(u)`

`\left(u+2\right)^3`

`=u^3+3u^2\cdot \:2+3u\cdot \:2^2+2^3`

`=u^3+6u^2+12u+8`

`=(u^3+6u^2+12u+8)/(u)`

`(u^3+6u^2+12u+8)/(u)=(u^3)/(u)+(6u^2)/(u)+(12u)/(u)+(8)/(u)`

`=u^2+(6u^2)/(u)+(12u)/(u)+(8)/(u)`

`=u^2+6u+(12u)/(u)+(8)/(u)`

`=u^2+6u+12+(8)/(u)`

`=\int \:u^2+6u+12+(8)/(u)du`

`=\int \:u^2du+\int \:6udu+\int \:12du+\int (8)/(u)du`

`\int \:u^2du`

`=(u^(2+1))/(2+1)`

`=(u^3)/(3)`

`\int \:6udu`

`=6\cdot \int \:udu`

`=6\cdot (u^(1+1))/(1+1)`

`=3u^2`

`\int \:12du`

`=12u`

`\int (8)/(u)du`

`=8\cdot \int (1)/(u)du`

`=8\ln \left|u\right|`

`=(u^3)/(3)+3u^2+12u+8\ln \left|u\right|`

`=(\left(x-2\right)^3)/(3)+3\left(x-2\right)^2+12\left(x-2\right)+8\ln \left|x-2\right|`

`=3x^2+8\ln \left|x-2\right|+(1)/(3)\left(x-2\right)^3-12`

`\int (2x^2)/(x-2)dx`

`=2\cdot \int (x^2)/(x-2)dx`

`=2\cdot \int (\left(u+2\right)^2)/(u)du`

`\left(u+2\right)^2`

`=u^2+4u+4`

`=(u^2+4u+4)/(u)`

`(u^2+4u+4)/(u)=(u^2)/(u)+(4u)/(u)+(4)/(u)`

`=u+(4u)/(u)+(4)/(u)`

`=2\cdot \int \:u+4+(4)/(u)du`

`=2\left(\int \:udu+\int \:4du+\int (4)/(u)du\right)`

`\int \:udu`

`=(u^(1+1))/(1+1)`

`=(u^2)/(2)`

`\int \:4du`

`=4u`

`\int (4)/(u)du`

`=4\cdot \int (1)/(u)du`

`=4\ln \left|u\right|`

`=2\left((u^2)/(2)+4u+4\ln \left|u\right|\right)`

`=2\left((\left(x-2\right)^2)/(2)+4\left(x-2\right)+4\ln \left|x-2\right|\right)`

`2\left((\left(x-2\right)^2)/(2)+4\left(x-2\right)+4\ln \left|x-2\right|\right)`

`=2\cdot (\left(x-2\right)^2)/(2)+2\cdot \:4\left(x-2\right)+2\cdot \:4\ln \left|x-2\right|`

`=x^2+4x+8\ln \left|x-2\right|-12`

`\int (3x)/(x-2)dx`

`=3\cdot \int (x)/(x-2)dx`

`=3\cdot \int (u+2)/(u)du`

`=3\cdot \int \:1+(2)/(u)du`

`=3\left(\int \:1du+\int (2)/(u)du\right)`

`\int \:1du`

`=u`

`\int (2)/(u)du`

`=2\cdot \int (1)/(u)du`

`=2\ln \left|u\right|`

`=3\left(u+2\ln \left|u\right|\right)`

`=3\left(x-2+2\ln \left|x-2\right|\right)`

`\int (1)/(x-2)dx`

`=\int (1)/(u)du`

`=\ln \left|u\right|`

`=\ln \left|x-2\right|`

`3x^2+8\ln \left|x-2\right|+(1)/(3)\left(x-2\right)^3-12-\left(x^2+4x+8\ln \left|x-2\right|-12\right)\\+3\left(x-2+2\ln \left|x-2\right|\right)-\ln \left|x-2\right|`

`=2x^2-x+5\ln \left|x-2\right|+(1)/(3)\left(x-2\right)^3-6`

`=2x^2-x+5\ln \left|x-2\right|+(1)/(3)\left(x-2\right)^3-6+C`

Answer 2

(⅓)x³ + 3x + 5ln|x - 2| + c

Explanation:

(x³ - 2x² + 3x - 6 + 5)/(x - 2)

[x²(x - 2) + 3(x - 2) + 5]/(x - 2)

x² + 3 + 5/(x - 2)

Integral:

⅓x³ + 3x + 5ln|x - 2| + c