integration using substitution

**cos16mh** · 13 Apr 2018, 11:44 AM

Hi all,

Could I have help with

∫(limits 1 and -3) dx/√(x+3) using x=u^2-3

**Drongoski** · 13 Apr 2018, 2:50 PM

Originally Posted by cos16mh

Hi all,

Could I have help with

∫(limits 1 and -3) dx/√(x+3) using x=u^2-3

$\int _{-3} ^1 \frac {1}{\sqrt {x+3}} dx = \int _{-3} ^1 (x+3)^{\frac {-1}{2}} d(x+3)$
$= \left [ \frac {(x+3)^{\frac {-1}{2}+1} }{\frac{1}{2}} \right ]^1 _{-3} = \left [2 \sqrt{x+3} \right] ^1 _{-3} = 4$

Why not use the Drongoski Method (if question does not insist on the substitution method!)?

**fan96** · 13 Apr 2018, 2:53 PM

$\int^{x=1}_{x=-3} \frac{1}{\sqrt{x+3}} \, dx$

Let $x = u^2-3$ , then $dx = 2u\, du$

The integral becomes

$\int^{u=2}_{u=0} \frac{2u}{\sqrt{u^2}} \, du$

(you can choose either $u=2$ or $u=-2$ , it doesn't change the answer)

$= \int^{u=2}_{u=0} \frac{2u}{|u|} \, du$

$= \int^{u=2}_{u=0} \frac{2u}{u} \, du$ ( $u \geq 0$ )

and the integral should be simple from there...

But another way is to recognise that we can integrate this by inspection:

$\frac{1}{\sqrt{x+3}} = (x+3)^{-\frac 1 2}$

$\int^1_{-3} (x+3)^{-\frac 1 2} \, dx = \left[2(x+3)^{\frac 1 2}\right]^1_{-3}$

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