Solveeit Logo
Login

Question

Mathematics Question on General and Particular Solutions of a Differential Equation

The integral of x2−xx3−x2+x−1\frac{x^{2}-x}{x^{3}-x^{2}+x-1} w.r.t. x is

A

12log(x2+1+c)\frac{1}{2}log \left(x^{2}+1+c\right)

B

12log∣x2−1∣+c\frac{1}{2}log \left|x^{2}-1\right|+c

C

log(x2+1+c)log \left(x^{2}+1+c\right)

D

log∣x2−1∣+clog \left|x^{2}-1\right|+c

Answer

12log(x2+1+c)\frac{1}{2}log \left(x^{2}+1+c\right)

Explanation

Solution

Let I=∫x2−xx2−x2+x−1dxI=\int \frac{x^{2 }-x}{x^{2} - x^{2} + x -1}dx =∫x(x−1)x2(x−1)+(x−1)dx=∫x dxx2+1= \int \frac{x\left(x-1\right)}{x^{2}\left(x-1\right) + \left(x-1\right)}dx = \int \frac{x\,dx}{x^{2}+1} =12∫2x dx(x2+1)= \frac{1}{2}\int\frac{2x\,dx}{\left(x^{2}+1\right)} Let x2+1=t⇒2x dx=dtx^{2} + 1 = t \Rightarrow 2x\, dx = dt ∴I=12∫dtt=12log t+c\therefore I = \frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}log \,t + c =12log(x2+1)+c= \frac{1}{2}log \left(x^{2}+1\right)+c where 'c' is the constant of integration.