Concept:

• \(\rm \int\frac{1}{x+a}dx=log|x+a|+C\)
• \(\rm \int\frac{1}{(x^2+1)}dx=tan^{-1}x+C\)

Calculation:​

\(\rm \int \frac{x^2-x+1}{(x+1)(x^2+1)}dx\)

This integrand is a proper rational fraction therefore, by using the form of partial fraction, we can write the integrand as

⇒\(\rm \frac{x^2-x+1}{(x+1)(x^2+2)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}\)

⇒ x2 - x + 1 = A(x² + 1) + (Bx + C) × (x + 1)

⇒ x2 - x + 1 = (A + B)x2 + (B + C)x + (A + C)

Comparing coefficient of x2, x and constant terms on both sides, we get A + B = 1, B + C = -1 and A + C = 1  

By solving these equation, we get  A = 3/2,  B = -1/2 and C = -1/2

⇒ \(\rm \frac{x^2-x+1}{(x+1)(x^2+2)}=\frac{3/2}{x+1}-\frac{1}{2}\frac{x+1}{(x^2+1)}\)  We can re-write it as 

⇒   \(\rm \frac{x^2-x+1}{(x+1)(x^2+2)}=\frac{3/2}{x+1}-\frac{1}{4}\frac{2x}{(x^2+1)}-\frac{1}{2}\frac{1}{(x^2+1)}\)

⇒\(\rm ​​\int\frac{x^2-x+1}{(x+1)(x^2+2)}dx=\frac{3}{2}\int\frac{1}{x+1}dx-\frac{1}{4}\int\frac{2x}{(x^2+1)}dx-\frac{1}{2}\int\frac{1}{(x^2+1)}dx\)

⇒\(\rm \int\frac{x^3-x+1}{(x+1)(x^2+1)}dx=\frac{3}{2}log\left | x+1 \right |-\frac{1}{4}log\left | x+1 \right |-\frac{1}{2}tan^{-1}x+C\)