Question

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

Answer 1

`(x^3+2x^2+5x-2)/((x^2+2x+2)^2)=(3x-2)/((x^2+2x+2)^2)+\frac x{x^2+2x+2}`

For the remaining integrals, we can complete the square in the denominator:


`x^2+2x+2=(x+1)^2+1`

and write each numerator as a polynomial
`x+1`:


`3x-2=3(x+1)-5`

`x=(x+1)-1`

then substitute
`x+1=\tan y`. Then
`\mathrm dx=\sec^2y\,\mathrm dy`, and we have


`\displaystyle\int\left((3\tan t-5)/((\tan^2t+1)^2)+(\tan t-1)/(\tan^2t+1)\right)\sec^2t\,\mathrm dt`

`\displaystyle\int\left((3\tan t-5)/(\sec^2t)+\tan t-1\right)\,\mathrm dt`

`\displaystyle\int(3\tan t\cos^2t-5\cos^2t+\tan t-1)\,\mathrm dt`

`=-\frac32\cos^2t-5\left(\frac12t+\frac14\sin2t\right)-\ln|\cos t|-t+C`

`=-\frac32\cos^2t-\frac54\sin2t-\frac72t-\ln|\cos t|+C`

Now
`t=\arctan(x+1)`, so the above simplifies to


`=-\frac3{2(x^2+2x+2)}-(5(x+1))/(2(x^2+2x+2))-\frac72\arctan(x+1)-\ln\frac1{√(x^2+2x+2)}+C`

`=-(5x+8)/(2(x^2+2x+2))-\frac72\arctan(x+1)+\frac12\ln(x^2+2x+2)+C`