Question

How do you evaluate the integral `int (xdx)/(x^2+4x+5)`?

Answer 1

`int \ x/(x^2+4x+5) \ dx = 1/2ln | x^2+4x+5 | - 2tan^-1 (x+2) + C`

Explanation:

Let

`I = int \ x/(x^2+4x+5) \ dx`

We can complete the square on the denominator, to get

`I = int \ x/((x+2)^2-2^2+5) \ dx`
`\ \ = int \ x/((x+2)^2+1) \ dx`

Let `u=x+2 iff x = u-2 => (dx)/(du) = 1`, so then substituting into the integral we get:

`I = int \ (u-2)/(u^2+1) \ du`
`\ \ = int \ u/(u^2+1) \ - 2/(u^2+1) \ du`
`\ \ = int \ u/(u^2+1) \ du \ - 2int 1 /(u^2+1) \ du`
`\ \ = 1/2int \ (2u)/(u^2+1) \ du \ - 2int 1 /(u^2+1) \ du`

The first intergral is of the form `int \ (f'(u))/(f(u)) \ du` which is a standard result with solution `ln | f(u) |`, And so:

`int \ (2u)/(u^2+1) \ du = ln | u^2+1 |`
`" "= ln | (x+2)^2+1 |`
`" "= ln | x^2+4x+5 |`

For the second integral, we can use the substitution `x=tan theta`

`u = tan theta => (du)/(d theta) = sec^2 theta`, and, `theta = tan^-1u`
`u = tan theta iff u^2=tan^2 theta`
`" " => 1+u^2=1+tan^2 theta`
`" " => 1+u^2=sec^2 theta`

Substituting into the integral we get:

`int 1 /(u^2+1) \ du = int \ d theta`
`" "= theta`
`" "= tan^-1u`
`" "= tan^-1 (x+2)`

Combining the result from both result we get:

`I = 1/2ln | x^2+4x+5 | - 2tan^-1 (x+2) + C`