Question

Integrate the following function: $\frac{1}{(x^2+1)(x^2+4)}$

Answer 1

Answer:

Step-by-step explanation:

Concept:

$\int{\frac{1}{1+x^2}}\:dx=tan^{-1}x+c\\\\\int{\frac{1}{x^2+a^2}}\:dx=\frac{1}{a}tan^{-1}(\frac{x}{a})+c$

The given integral is solved by decomposition method.

In decomposition method, the given integrand(non-integrable function) is decomposed into integrable functions by using algebraic identities, trigonometry identities, etc.

Now,

$\int{\frac{1}{(x^2+1)(x^2+4)}}\:dx\\\\=\int\frac{1}{3}(\frac{1}{x^2+1}-\frac{1}{x^2+4})\:dx\\\\=\frac{1}{3}\int(\frac{1}{x^2+1}-\frac{1}{x^2+4})\:dx\\\\=\frac{1}{3}[\int\frac{1}{x^2+1}\:dx-\int\frac{1}{x^2+^2}\:dx]\\\\=\frac{1}{3}[\int\frac{1}{1+x^2}\:dx-\int\frac{1}{x^2+^2}\:dx]\\\\=\frac{1}{3}[tan^{-1}x-\frac{1}{2}tan^{-1}(\frac{x}{2})]+c\\\\=\frac{1}{3}tan^{-1}x-\frac{1}{6}tan^{-1}(\frac{x}{2})+c$