## University Calculus: Early Transcendentals (3rd Edition)

$\frac{\theta^{2-\pi}}{2-\pi}+C$
Calculate the anti-derivative. we know that: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Then $\int \theta^{1-\pi} d\theta=\dfrac{\theta^{1-\pi+1}}{1-\pi+1}+C=\frac{\theta^{2-\pi}}{2-\pi}+C$