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

Question

Ezra Adams · Accepted Answer

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

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#

Cameron Alexander · Answer

undefined To evaluate the integral ∫ dx/(x^4+x^2), you can use the method of partial fractions. First, factor the denominator as x^2(x^2+1). Then, write the fraction as A/x + B/x^2 + C/(x^2+1). Find the values of A, B, and C using the method of equating coefficients. Once you have A, B, and C, integrate each term separately to get the final result.