Integrals Involving Quadratics — Question 7

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Question 7

Evaluate ∫x2x2+4dx\int\frac{x^2}{x^2+4}\,dx without partial fractions.

Original worksheet page 1: question and worked solution for 1-6-007
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Question 7 – Solution

Step 1: Rewrite the numerator. x2=(x2+4)−4.x^2=(x^2+4)-4. Therefore, x2x2+4=(x2+4)−4x2+4=1−4x2+4.\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=∫1dx−4∫dxx2+22.I=\int1\,dx-4\int\frac{dx}{x^2+2^2}. Step 3: Integrate both terms. I=x−4[12arctan(x2)]+C=x−2arctan⁡(x2)+C.\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*} x−2arctan⁡x2+C\boxed{x-2\arctan\frac{x}{2}+C}

Original worksheet page 2: question and worked solution for 1-6-007

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