(a)OR(b)

Find : $\displaystyle \int \frac{x^{3}+x}{x^{4}-9} \mathrm{~d} x$.

$$I = \int \frac{x^3 + x}{x^4 - 9}\,dx = \int \frac{x^3}{x^4 - 9}\,dx + \int \frac{x}{x^4 - 9}\,dx = \frac{1}{4}\log|x^4 - 9| + \frac{1}{2}\int \frac{dt}{t^2 - 3^2}\ ,\ \text{where } x^2 = t = \frac{1}{4}\log|x^4 - 9| + \frac{1}{2}\left[\frac{1}{2(3)}\log\left|\frac{t-3}{t+3}\right|\right] + C = \frac{1}{4}\log|x^4 - 9| + \frac{1}{12}\log\left|\frac{x^2 - 3}{x^2 + 3}\right| + C **Or** I = \int_{-\pi}^{\pi} (9\sin^2 x - 12\sin x + 4)\,dx = \int_{-\pi}^{\pi} 9\sin^2 x\,dx - \int_{-\pi}^{\pi} 12\sin x\,dx + \int_{-\pi}^{\pi} 4\,dx = 2\int_{0}^{\pi} 9\sin^2 x\,dx - 0 + 2\int_{0}^{\pi} 4\,dx (As $\displaystyle 9\sin^2 x$ and $\displaystyle 4$ are even and $\displaystyle 12\sin x$ is odd) = 9\int_{0}^{\pi} (1 - \cos 2x)\,dx + 8x\Big|_{0}^{\pi} = 9\left(x - \frac{\sin 2x}{2}\right)\Bigg|_{0}^{\pi} + 8\pi = 9\pi + 8\pi = 17\pi$$