Science:Math Exam Resources/Courses/MATH101/April 2015/Question 04 (b)/Solution 1

Using partial fraction decomposition, we can write the integrand as

\frac{1}{x^{4} + x^{2}} = \frac{1}{x^{2} (x^{2} + 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C x + D}{x^{2} + 1}

for some constants $A, B, C, D$ . We find that $A = 0, B = 1, C = 0, D = - 1$ , which implies that

\int \frac{1}{x^{4} + x^{2}} d x = \int \frac{1}{x^{2}} d x - \int \frac{1}{1 + x^{2}} d x .

The first integral on the right-hand side of the equation can be easily evaluated:

\int \frac{1}{x^{2}} d x = - \frac{1}{x} + C .

On the other hand, to compute the second integral, we can use the substitution $x = \tan u$ . Then $1 + x^{2} = 1 + \tan^{2} u = \sec^{2} u$ and $d x = \sec^{2} u d u$ , which gives

\int \frac{1}{1 + x^{2}} d x = \int \frac{1}{\sec^{2} u} \sec^{2} u d u = \int 1 d u = u + C = \arctan x + C .

Combining these, we obtain

\int \frac{1}{x^{4} + x^{2}} d x = - \frac{1}{x} - \arctan x + C .