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 P: 14 So this gives me $\int\frac{dv}{mg-kv^{2}}$ = $\int\frac{dt}{m}$ with u substitution: $\sqrt{\frac{k}{mg}v}$($\frac{1}{2}$$\int\frac{du}{1+u}$ + $\frac{1}{2}$$\int\frac{du}{1-u}$) = $\int\frac{dt}{m}$ becomes $\sqrt{\frac{k}{mg}v}(\frac{1}{2}$ln$\left|1+u\right|$ + $\frac{1}{2}$ln$\left|1-u\right|$) = $\frac{t}{m}$+c then $\sqrt{\frac{k}{mg}v}(ln\sqrt{1+u}$+ln$\sqrt{1-u}$)= $\frac{t}{m}$+c $\sqrt{\frac{k}{mg}v}(ln\sqrt{1-u^{2}}$)=$\frac{t}{m}$+c and finally sub back in for u to get: $\sqrt{\frac{k}{mg}v}*ln\sqrt{1-\frac{k}{mg}v^{2}}$=$\frac{t}{m}$+c Is this right?