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

$$\int\frac{e^{4t}+2e^{2t}-e^t}{e^{2t}+1}dt=\frac{e^{2t}}{2}+\frac{1}{2}\ln(e^{2t}+1)-\tan^{-1}e^t+C$$
$$I=\int\frac{e^{4t}+2e^{2t}-e^t}{e^{2t}+1}dt=\int\frac{e^{3t}+2e^t-1}{e^{2t}+1}(e^tdt)$$ Set $u=e^t$, which means $$du=e^tdt$$ We then have, $$I=\int\frac{u^3+2u-1}{u^2+1}du$$ We perform long division to get a polynomial and a proper fraction, which results in $$\frac{u^3+2u-1}{u^2+1}=u+\frac{u-1}{u^2+1}$$ Therefore, $$I=\int udu+\int\frac{u}{u^2+1}du-\int\frac{1}{u^2+1}du$$ $$I=\frac{u^2}{2}+\frac{1}{2}\ln|u^2+1|-\tan^{-1}u+C$$ $$I=\frac{u^2}{2}+\frac{1}{2}\ln(u^2+1)-\tan^{-1}u+C$$ (since $u^2+1\gt0$ for all $u$) $$I=\frac{e^{2t}}{2}+\frac{1}{2}\ln(e^{2t}+1)-\tan^{-1}e^t+C$$