How do you integrate #int (3x) / (x^2 * (x^2+1) )# using partial fractions?

The expression needs some simplification

#(3x)/(x^2(x^2+1))=3/(x(x^2+1))#

Now we can perform the decomposition into partial fractions

#1/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)#

#=(A(x^2+1)+x(Bx+C))/(x(x^2+1))#

The denominators are the same, we compare the numerators

#1=(A(x^2+1)+x(Bx+C))#

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

Coefficients of #x^2#,

#0=A+B#

#B=-1#

Coefficients of #x#,

#0=C#

Therefore,

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

So,

the integral is

#3int(1dx)/(x(x^2+1))=3intdx/x-3int(xdx)/(x^2+1)#

The second integral is done by substutution

Let #u=x^2+1#

#du=2xdx#

#int(xdx)/(x^2+1)=1/2int(du)/u#

#=1/2lnu#

#=1/2ln(x^2+1)#

Putting it all together,

#3int(1dx)/(x(x^2+1))=3ln(|x|)-3/2ln(x^2+1)+C#

Resprcted Narad T. has solved the Problem using Partial

Fractions, as is reqd. to be done in that fashion.

As an Aliter, I submit the following Soln. :

Let #I=int(3x)/{x^2(x^2+1)}dx=3/2int(2x)/{x^2(x^2+1)}dx.#

Noting that, #d/dx(x^2)=2x,# we subst. #x^2=t," so, "2xdx=dt.#

#:. I=3/2int1/{t(t+1)}dt=3/2int{(t+1)-(t)}/{t(t+1)}dt,#

#=3/2int{(t+1)/(t(t+1))-t/(t(t+1))}dt,#

#=3/2int{1/t-1/(t+1)}dt,#

#=3/2[ln|t|-ln|t+1|},#

#=3/2ln|t/(t+1)|, and, because, t=x^2,#

#I=3/2ln{x^2/(x^2+1)}+C, or, #

#I=3/2lnx^2-3/2ln(x^2+1)+C=3ln|x|-3/2ln(x^2+1)+C.#

Enjoy Maths.!