Calculus simplification

Question

Could someone explain how this:

$\int(\frac{x}{\sqrt{x+1}})dx$

becomes:

$\int(\sqrt{x+1})dx - \int(\frac{1}{\sqrt{x+1}})dx$

Answer 1

$$\int(\sqrt{x+1})dx - \int(\frac{1}{\sqrt{x+1}})dx$$ $$\int(\sqrt{x+1}) -\frac{1}{\sqrt{x+1}}dx$$

multiply both numerator and denominator by the same value $\sqrt {(x+1)}$

$$\int \frac {(x+1)-1}{\sqrt{x+1}}dx$$ $$\int \frac x{\sqrt{x+1}}dx$$

Answer 2

This does not come from integration by parts, but from substitution. Let $u=x+1$ so that $du=dx$ and $x=u-1$. Then

$$ \int \frac{x}{\sqrt{x+1}} \ dx = \int \frac{u-1}{\sqrt{u}} \ du = \int \sqrt{u} - \frac{1}{\sqrt{u}} \ du. $$ The last integral is $$ \int \sqrt{x+1} - \frac{1}{\sqrt{x+1}} \ dx $$ which is what you claim. However, it is more easily done from the $u$ line since $$ \int \sqrt{u} - \frac{1}{\sqrt{u}} \ du = \frac{2}{3}u^{3/2} - 2\sqrt{u} + C. $$ Sub back in $u=x+1$ and you're done.

Answer 3

It’s got nothing to do with the integral:

$$\begin{align} \frac{x}{\sqrt{x+1}} &= \frac{x+1-1}{\sqrt{x+1}}\\ &= \frac{x+1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}}\\ &= \sqrt{x+1} - \frac{1}{\sqrt{x+1}} \end{align}$$

Does that help?