Integrate `int(x+4)/(x^2+2x+5)dx`

`int(x+4)/(x^2+2x+5)dx=int(x+1)/(x^2+2x+5)dx+int3/(x^2+2x+5)dx`

Integrate the first integral on the left side of the equation using the u-substitution method.

Let `u=x^2+2x+5`

`(du)/(dx)=2x+2`

`dx=(du)/(2(x+1))`

`int(x+1)/(x^2+2x+5)dx`

`=int(x+1)/u*(du)/(2(x+1))`

`=1/2ln|u|+C`

`1/2ln|x^2+2x+5|+C`

The second integral on the left side will match the form

`intdx/(x^2+a^2)=1/atan^-1(x/a)+C` after you complete the square in the denominator.

`int3/(x^2+2x+5)dx`

`=3intdx/[(x^2+2x+1)+5-1]`

`=3intdx/[(x+1)^2+2^2]`

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

`=(3/2)tan^-1[(x+1)/2]+C`

**The final answer is:**

`1/2ln|x^2+2x+5|+(3/2)tan^-1[(x+1)/2]+C`

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