Integral x^2 sqrt (x+1) dx

Question

asked Jul 11, 2019 23.0k views

1 Answer

M Newville · Answer 1 · 2019-07-15T11:14:32+0000

If you mean to find the antiderivative

$\displaystyle\int x^2√(x+1)\,\mathrm dx$

substitute
y=x+1 , so that
x=y-1 and
$\mathrm dx=\mathrm dy$ . Then expand the squared term and distribute
$\sqrt y$ :

$\displaystyle\int(y-1)^2\sqrt y\,\mathrm dy=\int\left(y^(5/2)-2y^(3/2)+y^(1/2)\right)\,\mathrm dy$

$=\frac27y^(7/2)-\frac45y^(5/2)+\frac23y^(3/2)+C$

$=\frac2{105}y^(3/2)\left(15y^2-42y+35\right)+C$

$=\frac2{105}(x+1)^(3/2)\left(15(x+1)^2-42(x+1)+35\right)+C$

$=\frac2{105}(x+1)^(3/2)\left(15x^2-12x+8\right)+C$

Or if instead you mean to write

$\displaystyle\int(x^2)/(√(x+1))\,\mathrm dx$

the same substitution as before leads to

$\displaystyle\int((y-1)^2)/(\sqrt y)\,\mathrm dy=\int\left(y^(3/2)-2y^(1/2)+y^(-1/2)\right)\,\mathrm dy$

$=\frac25y^(5/2)-\frac43y^(3/2)+2y^(1/2)+C$

$=\frac2{15}y^(1/2)\left(3y^2-10y+15\right)+C$

$=\frac2{15}(x+1)^(1/2)\left(3(x+1)^2-10(x+1)+15\right)+C$

$=\frac2{15}(x+1)^(1/2)\left(3x^2-4x+8\right)+C$