Answer
$\displaystyle \int (1+\sec(\pi x))^2\sec(\pi x)\tan(\pi x)dx = \dfrac{(1+\sec(\pi x))^3}{3\pi} + K $
Work Step by Step
$u = 1+\sec(\pi x) \quad \rightarrow \quad du = [1+\sec(\pi x)]'dx = \pi \sec(\pi x)\tan(\pi x)dx \quad \\ \rightarrow \dfrac{du}{\pi} = \sec(\pi x)\tan(\pi x)dx$
$\displaystyle \int (1+\sec(\pi x))^2\sec(\pi x)\tan(\pi x)dx = \displaystyle \int u^2\dfrac{du}{\pi} = \displaystyle \dfrac{1}{\pi}\int u^2du = \dfrac{1}{\pi}\bigg[\dfrac{u^3}{3} + C\bigg] = \dfrac{u^3}{3\pi} + \dfrac{C}{\pi} = \dfrac{(1+\sec(\pi x))^3}{3\pi} + K $
