Chapter 8: Techniques of Integration - Section 8.6 - Integral Tables and Computer Algebra Systems - Exercises 8.6 - Page 481: 6

$\dfrac{(7x+5)^{5/2}}{49}[\dfrac{14x-4}{7}]+C$

By using Formula 22, we get: $\int x(ax+b)^ndx=\dfrac{(ax+b)^{n+1}}{a^2} \space [\dfrac{ax+b}{n+2}-\dfrac{b}{n+1}]+ \space C$ Let $I =\dfrac{(7x+5)^{5/2}}{7^2}[\dfrac{7x+5}{\dfrac{7}{2}}-\dfrac{5}{\dfrac{5}{2}}\Big]+C \\=\dfrac{(7x+5)^{5/2}}{49} [\dfrac{14x+10}{7}-2]+C \\=\dfrac{(7x+5)^{5/2}}{49}[\dfrac{14x-4}{7}]+C$