Question

Integral of x(2x+3)^99 dx

Answer 1

Let
`y=2x+3`, so that
`x=\frac{y-3}2` and so
`\mathrm dx=\frac{\mathrm dy}2`. Then the integral is


`\displaystyle\int x(2x+3)^(99)\,\mathrm dx=\int\left(\frac{y-3}2\right)y^(99)\frac{\mathrm dy}2`

`=\displaystyle\frac14\int(y^(100)-3y^(99))\,\mathrm dy`

`=\displaystyle(y^(101))/(404)-(3y^(100))/(400)+C`

`=\displaystyle((2x+3)^(101))/(404)-(3(2x+3)^(100))/(100)+C`

`=((2x+3)^(100))/(40400)(100(2x+3)-303)+C`

`=((2x+3)^(100))/(40400)(200x-3)+C`