# Integration using Substitution

• Jan 23rd 2010, 01:31 PM
seuzy13
Integration using Substitution
1.
$\int\frac{-x}{(x + 1) - \sqrt{x + 1}} dx$

2.
$\int^2_1(x - 1)\sqrt{2 - x} dx$

I am stumped on both of these problems.
On #1, I try u = x + 1, but wind up with
$\int\frac{-u + 1}{u - u^\frac{1}{2}} du$
which only seems more unsolvable than it was before

On #2, I try u = 2 - x, but then wind up getting 1 for the lower limit and 0 for the upper limit which makes no sense.

• Jan 23rd 2010, 01:39 PM
pickslides
Quote:

Originally Posted by seuzy13
1.

$\int\frac{-u + 1}{u - u^\frac{1}{2}} du$
which only seems more unsolvable than it was before

This might help.

$\int\frac{-u }{u - u^\frac{1}{2}} du+\int\frac{ 1}{u - u^\frac{1}{2}} du$

Now factor out u and cancel in the first integral.
• Jan 23rd 2010, 01:41 PM
Krizalid
Quote:

Originally Posted by seuzy13
On #2, I try u = 2 - x, but then wind up getting 1 for the lower limit and 0 for the upper limit which makes no sense.

That does make sense. Consider $\int_a^b f=g(b)-g(a)=-\big(g(a)-g(b)\big)=-\int_b^a f,$ so if you flip the bounds, then the integral changes its sign.
• Jan 23rd 2010, 01:43 PM
seuzy13
Hmm...but now I'm winding up with:
$\int\frac{-1}{1 - u^\frac{-1}{2}} + \frac{1}{u} - \frac{1}{u^\frac{1}{2}} du$
In which I the first term still seems unsolvable...?

EDIT--
Oh and thanks for the clarification on #2. I had forgotten about that rule.
• Jan 23rd 2010, 03:27 PM
$\int{\frac{-x}{x+1-\sqrt{x+1}}}dx=-\int{\frac{x}{\sqrt{x+1}^2-\sqrt{x+1}}}dx=-\int{\frac{u^2-1}{u^2-u}}2udu$

where $u=\sqrt{x+1}$

so $x=u^2-1$

giving

$-\int{\frac{(u+1)(u-1)}{u(u-1)}}2udu=-2\int{(u+1)}du=-2(\frac{u^2}{2}+u)+C$

$=-2\ \frac{x+1}{2}-2\sqrt{x+1}+C$

• Jan 23rd 2010, 03:52 PM
drumist

I think this is the easiest method:

Let $u=x+1 \implies x=u-1$

$
\int \frac{-x}{x+1-\sqrt{x+1}} dx
=\int \frac{-(u-1)}{u-u^{1/2}}du
=\int \frac{-(u^{1/2}+1)(u^{1/2}-1)}{u^{1/2}(u^{1/2}-1)}du
$

$
=\int \frac{-(u^{1/2}+1)}{u^{1/2}}du
=\int \left(-1 - u^{-1/2}\right) du
= -u - 2u^{1/2} = ...$