Question

How do I find the integral `int(x^3+4)/(x^2+4)dx` ?

Answer 1

`I=1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+c`, where `c` is a constant

Explanation,

`I=int(x^3+4)/(x^2+4)dx`

`I=int((x^3)/(x^2+4)+4/(x^2+4))dx`

`I=int(x^3)/(x^2+4)dx+int4/(x^2+4)dx`

`I=I_1+I_2` .......`(i)`

Now considering only first integral, which is `I_1=int(x^3)/(x^2+4)dx`

Using Integration by Substitution,

let's `x^2=t` , then `2xdx=dt` yields, first integral

`=intt/2*1/(t+4)*dt`

`=1/2intt/(t+4)*dt`, this can be written as

`=1/2int(t+4-4)/(t+4)*dt`

`=1/2int(1-4/(t+4))dt`

`=1/2intdt-2int1/(t+4)dt`

`=1/2t-2ln(t+4)+c_1`, where `c_1` is a constant

replacing `t`, we get,

`I_1=1/2x^2-2ln(x^2+4)+c_1`, where `c_1` is a constant

considering second integral

`I_2=int4/(x^2+4)dx`

using Trigonometric Substitution to solve this problem,

`I_2=4*1/2tan^-1(x/2)+c_2`

`I_2=2tan^-1(x/2)+c_2`

Finally, plugging in both `I_1` and `I_2` in `(i)`

`I=1/2x^2-2ln(x^2+4)+c_1+2tan^-1(x/2)+c_2`

`I=1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+c`, where `c=c_1+c_2` is again a constant