Substitution Method Help

**Vamz** · December 4th 2010, 11:17 PM

The substitution $u=2x+7$ transforms the "x"-integral
$\displaystyle \int \frac{4 x + 3}{\sqrt{2 x + 7}} \ dx$
into the "u"-integral

ok, so I can rewrite this as:

$\displaystyle \int (4 x + 3)(2 x + 7)^{\frac{-1}{2}} \ dx$
$u=2x+7$
$du=2dx$
I can't figure out how to force this to equal 4x+3

All I can do is
$2\int u^{\frac{-1}{2}}*du$

which will give me my 4, but I'm still missing the x+3 part!

What am I missing?

**Sudharaka** · December 4th 2010, 11:59 PM

Originally Posted by Vamz

The substitution $u=2x+7$ transforms the "x"-integral
$\displaystyle \int \frac{4 x + 3}{\sqrt{2 x + 7}} \ dx$
into the "u"-integral

ok, so I can rewrite this as:

$\displaystyle \int (4 x + 3)(2 x + 7)^{\frac{-1}{2}} \ dx$
$u=2x+7$
$du=2dx$
I can't figure out how to force this to equal 4x+3

All I can do is
$2\int u^{\frac{-1}{2}}*du$

which will give me my 4, but I'm still missing the x+3 part!

What am I missing?

Dear Vamz,

$\displaystyle \int \frac{4 x + 3}{\sqrt{2 x + 7}} \ dx$

$u=2x+7\Rightarrow{dx=\frac{du}{2}}$

Also, $x=\frac{u-7}{2}$

Substituting these in the above integration;

$\displaystyle \int \frac{4 \left(\frac{u-7}{2}\right) + 3}{2\sqrt{u}} \ du$

$\displaystyle\frac{1}{2}\int \frac{2u-11}{\sqrt{u}}du$

Hope you can continue.

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