AP Calculus: evaluate int [ x sqrt(x + 1) ] dx

[venialove]

Not sure if mine is right (can you show the steps please)

Evaluate the integral:

?x ?(x+1) dx

this is what i did:
u=x
du=?x+1

?x ?(x+1) dx =? x(x+1)^1/2 dx
(2/3)x (x+1)^3/2-(4/15)(x+1) 3/2+c

 

[soroban]

Re: ap calculus (can you show the steps please)

Hello, venialove!

Sorry, I can't follow what you did . . .

\(\displaystyle \int x \sqrt{x+1}\,dx\)

\(\displaystyle \text{Let }\,u \:=\:\sqrt{x+1}\quad\Rightarrow\quad x \:=\:u^2-1\quad\Rightarrow\quad dx \:=\:2u\,du\)

\(\displaystyle \text{Substitute: }\;\int(u^2-1)\!\cdot\!u\!\cdot\!2u\,du \;=\;2\int(u^4 - u^2)\,du\)

. . . \(\displaystyle =\;2\left(\frac{1}{5}u^5 - \frac{1}{3}u^3\right) + C \;=\;\frac{2}{15}u^3\left(3u^2-5\right) + C\)

\(\displaystyle \text{Back-substitute: }\;\frac{2}{15}\left(\sqrt{x+1}\right)^3\bigg[3(\sqrt{x+1})^2 - 5\bigg] + C\)

. . . . . . . . . . \(\displaystyle = \;\frac{2}{15}(x+1)^{\frac{3}{2}}\bigg[3(x+1) - 5\bigg] + C\)

. . . . . . . . . . \(\displaystyle = \;\frac{2}{5}(x+1)^{\frac{3}{2}}(3x + 3 - 5) + C\)

. . . . . . . . . . \(\displaystyle = \;\frac{2}{15}(x+1)^{\frac{3}{2}}(3x-2) + C\)