Calculus: Early Transcendentals 8th Edition

$\displaystyle\int\limits_1^2\dfrac{v^{3}+3v^{6}}{v^{4}}dv=\ln2+ 7$
$\displaystyle\int\limits_1^2\dfrac{v^{3}+3v^{6}}{v^{4}}dv$ Take the denominator up to multiply the numerator by changing the sign of its exponent: $\displaystyle\int\limits_1^2(v^{3}+3v^{6})v^{-4}dv=\int\limits_1^2(v^{-1}+3v^{2})dv=\int\limits_1^2\Big(\dfrac{1}{v}+3v^{2}\Big)dv=...$ Integrate each term separately and apply the second part of the fundamental theorem of calculus: $\displaystyle\int\limits_1^2\dfrac{1}{v}dv+3\int\limits_1^2v^{2}dv=\ln v+3\Big(\dfrac{1}{3}\Big)v^{3}\Big|_1^2=\ln v+v^{3}\Big|_1^2=...$ $...=[\ln2+(2)^{3}]-[\ln1+(1)^{3}]=\ln2+8-0-1=\ln2+ 7$