Partial Fractions — Question 6

Question 6 – Solution

Part 1: Check the decomposition.

Each power of an irreducible quadratic needs a linear numerator. The general form is $$\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}.$$ The original fraction already has this form: $A=B=C=0$ and $D=1$. Therefore, decomposing it only reproduces the original integrand and does not make the integral easier.

Part 2: Evaluate the integral.

Step 1: Substitute $x=\tan\theta$. Then $$dx=\sec^2\theta\,d\theta, \qquad x^2+1=\tan^2\theta+1=\sec^2\theta.$$ Step 2: Simplify and integrate. $$\begin{align*} I&=\int\frac{\sec^2\theta}{\sec^4\theta}\,d\theta =\int\cos^2\theta\,d\theta\\ &=\frac12\int(1+\cos2\theta)\,d\theta\\ &=\frac{\theta}{2}+\frac{\sin2\theta}{4}+C. \end{align*}$$ Step 3: Return to $x$. $$\theta=\arctan x,\qquad \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2x}{1+x^2}.$$ Substitute both expressions: $$\boxed{\displaystyle I=\frac{x}{2(x^2+1)} +\frac12\arctan x+C}.$$