Let [tex]y=2x+3[/tex], so that [tex]x=\dfrac{y-3}2[/tex] and so [tex]\mathrm dx=\dfrac{\mathrm dy}2[/tex]. Then the integral is
[tex]\displaystyle\int x(2x+3)^{99}\,\mathrm dx=\int\left(\frac{y-3}2\right)y^{99}\dfrac{\mathrm dy}2[/tex]
[tex]=\displaystyle\frac14\int(y^{100}-3y^{99})\,\mathrm dy[/tex]
[tex]=\displaystyle\frac{y^{101}}{404}-\frac{3y^{100}}{400}+C[/tex]
[tex]=\displaystyle\frac{(2x+3)^{101}}{404}-\frac{3(2x+3)^{100}}{100}+C[/tex]
[tex]=\dfrac{(2x+3)^{100}}{40400}(100(2x+3)-303)+C[/tex]
[tex]=\dfrac{(2x+3)^{100}}{40400}(200x-3)+C[/tex]
