Question

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

Answer 1

The answer is `=-1/x-arctanx+C`

Explanation:

We need

`intx^ndx=x^(n+1)/(n+1)+C(n!=-1)`

We factorise the denominator

`x^4+x^2=x^2(x^2+1)`

Now we can do the decomposition into partial fractions

`1/(x^4+x^2)=A/x^2+B/x+(Cx+D)/(x^2+1)`

`=(A(x^2+1)+Bx(x^2+1)+(Cx+D)(x^2))/(x^2(x^2+1))`

Therefore,

`1=A(x^2+1)+Bx(x^2+1)+(Cx+D)(x^2)`

Let, `x=0`,`=>`,`1=A`

Coefficients of `x^2`, `=>`, `0=A+D`, `=>`, `D=-1`

Coefficients of `x`, `=>`, `0=B`

Coeficients of `x^3`, `=>`, `0=B+C`, `=>`, `C=0`

So,

`1/(x^4+x^2)=1/x^2+0/x+(0x-1)/(x^2+1)`

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

Therefore,

`intdx/(x^4+x^2)=intdx/x^2-intdx/(x^2+1)`

The first integral is `intdx/x^2=-1/x`

The second integral is `intdx/(x^2+1)`

We use a trigonometric substitution

Let `x=tanu`, `=>`, `du=sec^2udu`

and `x^2+1=tan^2u+1=sec^2u`

So,

`intdx/(x^2+1)=int(sec^2udu)/sec^2u=intdu=u`

`=arctanx`

Putting it alltogether,

`intdx/(x^4+x^2)=-1/x-arctanx+C`