Question

Evaluate the integral? : `int x^2/(x^2+1)^2 dx`

(Question Restore: portions of this question have been edited or deleted!)

Answer 1

Option #1

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

Explanation:

Let us denote the given integral by `I`:

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

We can simply the fractional integrand with a slight manipulation:

`I = int \ (x^2+1-1)/(x^2+1)^2 \ dx`
`\ \ = int \ (x^2+1)/(x^2+1)^2 - (1)/(x^2+1)^2 \ dx`
`\ \ = int \ 1/(x^2+1) \ dx - int \ (1)/(x^2+1)^2 \ dx`
`\ \ = arctanx - int \ (1)/(x^2+1)^2 \ dx + C`

For this next integral we can use a substitution: Let

`tan theta = x => x^2 +1 = tan^2 theta+1 = sec^2 theta`
`dx/(d theta)=sec^2 theta`

Substituting we see that:

`int \ (1)/(x^2+1)^2 \ dx = int \ (1)/(sec^2 theta)^2\ sec^2 theta \ d theta`
`" " = int \ (1)/(sec^2 theta)^2\ sec^2 theta \ d theta`
`" " = int \ (1)/(sec^2 theta) \ d theta`
`" " = int \ cos^2 theta \ d theta`
`" " = int \ 1/2(1+cos2theta) \ d theta`
`" " = 1/2 \ int \ 1+cos2theta \ d theta`
`" " = 1/2 (theta + 1/2sin2theta)`
`" " = 1/2 (theta + sinthetacostheta)`
`" " = 1/2 (theta + sinthetacostheta * costheta/costheta)`
`" " = 1/2 (theta + sintheta/costheta cos^2theta)`
`" " = 1/2 (theta + tan theta/sec^2theta)`
`" " = 1/2 (arctanx + x/(x^2+1) )`

Combining this with the earlier result gives:

`I = arctanx - 1/2 (arctanx + x/(x^2+1) ) + C`
`\ \ = 1/2arctanx - x/(2(x^2+1)) + C`