Integrals Involving Quadratics — Question 7

Question 7

Evaluate $$\int\frac{x^2}{x^2+4}\,dx$$ without partial fractions.

Question 7 – Solution

Step 1: Rewrite the numerator. $$x^2=(x^2+4)-4.$$ Therefore, $$\begin{align*} \frac{x^2}{x^2+4} &=\frac{(x^2+4)-4}{x^2+4}\\ &=1-\frac4{x^2+4}. \end{align*}$$ Step 2: Split the integral. $$I=\int1\,dx-4\int\frac{dx}{x^2+2^2}.$$ Step 3: Integrate both terms. $$\begin{align*} I&=x-4\left[\frac12\arctan\left(\frac{x}{2}\right)\right]+C\\ &=x-2\arctan\left(\frac{x}{2}\right)+C. \end{align*}$$ $$\boxed{x-2\arctan\frac{x}{2}+C}$$