How do you integrate int (3x) / (x^2 * (x^2+1) )∫3xx2⋅(x2+1) using partial fractions?

2 Answers
Narad T.
Apr 5, 2017

The answer is =3ln(|x|)-3/2ln(x^2+1)+C=3ln(|x|)−32ln(x2+1)+C

Explanation:

The expression needs some simplification

(3x)/(x^2(x^2+1))=3/(x(x^2+1))3xx2(x2+1)=3x(x2+1)

Now we can perform the decomposition into partial fractions

1/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)1x(x2+1)=Ax+Bx+Cx2+1

=(A(x^2+1)+x(Bx+C))/(x(x^2+1))=A(x2+1)+x(Bx+C)x(x2+1)

The denominators are the same, we compare the numerators

1=(A(x^2+1)+x(Bx+C))1=(A(x2+1)+x(Bx+C))

Let x=0x=0, =>⇒, 1=A1=A

Coefficients of x^2x2,

0=A+B0=A+B

B=-1B=−1

Coefficients of xx,

0=C0=C

Therefore,

1/(x(x^2+1))=1/x-x/(x^2+1)1x(x2+1)=1x−xx2+1

So,

the integral is

3int(1dx)/(x(x^2+1))=3intdx/x-3int(xdx)/(x^2+1)3∫1dxx(x2+1)=3∫dxx−3∫xdxx2+1

The second integral is done by substutution

Let u=x^2+1u=x2+1

du=2xdxdu=2xdx

int(xdx)/(x^2+1)=1/2int(du)/u∫xdxx2+1=12∫duu

=1/2lnu=12lnu

=1/2ln(x^2+1)=12ln(x2+1)

Putting it all together,

3int(1dx)/(x(x^2+1))=3ln(|x|)-3/2ln(x^2+1)+C3∫1dxx(x2+1)=3ln(|x|)−32ln(x2+1)+C

Ratnaker Mehta
Apr 5, 2017

3/2ln{x^2/(x^2+1)}+C, or, 32ln{x2x2+1}+C,or,

3ln|x|-3/2ln(x^2+1)+C.3ln|x|−32ln(x2+1)+C.

Explanation:

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.I=∫3xx2(x2+1)dx=32∫2xx2(x2+1)dx.

Noting that, d/dx(x^2)=2x,ddx(x2)=2x, we subst. x^2=t," so, "2xdx=dt.x2=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.!

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
1608 views around the world
You can reuse this answer
Creative Commons License