Do you mean integrate ∫(1+3v2)dv/(v(1-v2))? The expression you have provided is not a differential equation. An equation would contain an equals sign and a differential equation would also contain at least one differential component such as dv/dx. You can always resubmit your question or amend it by editing.
I'll assume that's an integration.
(1+3v2)dv/(v(1-v2)) can be expressed in partial fractions which will usually simplify integration.
(1+3v2)/(v(1-v2))≡A/v+B/(1-v)+C/(1+v) because 1-v2=(1-v)(1+v). A, B, C are constants.
1+3v2≡A-Av2+Bv+Bv2+Cv-Cv2. By matching coefficients:
(1) constant: A=1;
(2) v2: -A+B-C=3;
(3) v: B+C=0.
Substitute A=1 in (2): -1+B-C=3, so B-C=4;
Add this to (3): 2B=4, B=2, therefore C=-2.
(1+3v2)/(v(1-v2))≡A/v+B/(1-v)+C/(1+v)=1/v+2/(1-v)-2/(1+v).
Now the integral becomes:
∫dv/v+2∫dv/(1-v)-2∫dv/(1+v)=ln(v)-2ln(1-v)-2ln(1+v)+C, where C is the integration constant.
Note that 2ln(x) is the same as ln(x2) where x is any quantity, and -ln(x) is the same as ln(1/x).
Therefore, -2[ln(1-v)+ln(1+v)]=-2ln[(1-v)(1+v)]=-2ln(1-v2)=-ln(1-v2)2=ln(1/(1-v2)2).
ln(v/(1-v2)2)+C is the solution.