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:​

To solve \(\rm \int \frac{2x^3}{(x^2+2)(x^4+1)}dx\) 

Let x² = t ⇒ 2x dx = dt

Let us put this in given integration, then our integration becomes

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

This integrand is a proper rational fraction. So, by using the form of partial fraction, we write

⇒\(\rm \frac{t}{(t+2)(t^2+1)}=\frac{A}{t+2}+\frac{Bt+C}{t^2+1}\)

⇒ t = A(t² + 2) + (Bt + C)(t + 2)

 ⇒ t = (A + B)t2 + (2B + C)t + 2(A + C)

Comparing coefficient of t2, t and constant terms on both sides, we get A + B = 0, 2B + C = 1 and A + 2C = 0  

By solving these equation, we get  A = -2/5,  B = 2/5 and C = 1/5

Thus  \(\rm \frac{t}{(t+2)(t^2+1)}=\frac{-2/5}{t+2}+\frac{1}{5}\frac{2t+1}{(t^2+1)}\) we can re-write it as 

⇒ \(\rm \frac{t}{(t+2)(t^2+1)}=-\frac{2}{5}\frac{1}{t+2}+\frac{1}{5}\frac{2t}{(t^2+1)}+\frac{1}{5}\frac{1}{(t^2+1)}\)

Therefore, \(\rm \int\frac{t}{(t+2)(t^2+1)}dt=-\frac{2}{5}\int\frac{1}{t+2}dt+\frac{1}{5}\int\frac{2t}{(t^2+1)}dt+\frac{1}{5}\int\frac{1}{(t^2+1)}dt\)

⇒ \(\rm \int\frac{t}{(t+2)(t^2+1)}dt=\frac{-2}{5}log\left | t+2 \right |+\frac{1}{5}log\left | t^2+1 \right |+\frac{1}{5}tan^{-1}t+C\) 

Now put t = x² in the above equation we get

⇒ \(\rm \int\frac{2x^3}{(x^2+2)(x^4+1)}dx=\frac{-2}{5}log\left | x^2+2 \right |+\frac{1}{5}log\left | x^4+1 \right |+\frac{1}{5}tan^{-1}x^2+C\)