integral of y/sqrt (y+1) dy

[math]

integral of y/sqrt (y+1) dy

Here's my work:

y = u-1 u=y+1
u= sqrt (y+1) v' = y
u' = 1/(2 sqrt(y+1)) v = y^2 / 2

Integration by Parts:
1/2 y^2 sqrt (y+1) - integral of 1/2 y^2 X 1/(2 sqrt (y+1))
1/2 y^2 sqrt (y+1) - integral y^2 / (4 sqrt(y+1))

any suggestions?

 

[soroban]

Hello, math!

I recommend straight substitution . . .

\(\displaystyle \L\int \frac{y}{\sqrt{y\,+\,1}}\,dy\)

Let \(\displaystyle u \:=\:\sqrt{y\,+\,1}\;\;\Rightarrow\;\;y \:=\:u^2\,-\,1\;\;\Rightarrow\;\;dy\:=\:2u\,du\)

Substitute: \(\displaystyle \L\:\int\frac{u^2\,-\,1}{u}\cdot(2u\,du) \;=\;2\int(u^2\,-\,1)\,du\)

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


Back-substitute: \(\displaystyle \L\:\frac{2}{3}\sqrt{y\,+\,1}\;\left[(\sqrt{y\,+\,1})^2\,-\,3\right]\,+\,C \;=\;\frac{2}{3}\sqrt{y\,+\,1}\;\left[y\,+\,1\,-\,3\right]\)


Answer: \(\displaystyle \L\;\frac{2}{3}\sqrt{y\,+\,1}\,(y\,-\,2)\,+\,C\)

 

[math]

thanks!!