## Calculus: Early Transcendentals 8th Edition

$\frac{32}{3}\ln 2-\frac{28}{9}$
$\int_1^4\sqrt{y}\ln y dy$ Use integration by parts with $u=\ln y$, $du=\frac{1}{y}dy$, $dv=\sqrt{y}dy=y^{1/2}dy$, $v=\frac{y^{3/2}}{\frac{3}{2}}=\frac{2y^{3/2}}{3}$. Since $\int_a^b u\ dv=uv|_a^b-\int_a^b v\ du$, the original integral is equal to: $=(\ln y*\frac{2y^{3/2}}{3})|_1^4-\int_1^4 \frac{2y^{3/2}}{3}*\frac{1}{y}dy$ $=(\ln 4*\frac{2*4^{3/2}}{3}-\ln 1*\frac{2*1^{3/2}}{3})-\int_1^4 \frac{2}{3}y^{1/2}dy$ $=(\ln 4*\frac{2*8}{3}-0)-(\frac{2}{3}*\frac{y^{3/2}}{\frac{3}{2}})|_1^4$ $=\frac{16}{3}\ln 4-(\frac{2}{3}*\frac{4^{3/2}}{\frac{3}{2}}-\frac{2}{3}*\frac{1^{3/2}}{\frac{3}{2}})$ $=\frac{16}{3}\ln 4-(\frac{2}{3}*\frac{8}{\frac{3}{2}}-\frac{2}{3}*\frac{1}{\frac{3}{2}})$ $=\frac{16}{3}\ln 4-(\frac{2}{3}*\frac{16}{3}-\frac{2}{3}*\frac{2}{3})$ $=\frac{16}{3}\ln 4-(\frac{32}{9}-\frac{4}{9})$ $=\frac{16}{3}\ln 4-\frac{28}{9}$ Using the logarithm rule $\ln a^b=b\ln a$, this can be rewritten as: $=\frac{16}{3}\ln (2^2)-\frac{28}{9}$ $=\frac{16}{3}*2\ln 2-\frac{28}{9}$ $=\boxed{\frac{32}{3}\ln 2-\frac{28}{9}}$