1. ## Simple Integration

$\int \frac{u}{\sqrt{u^2 +16}} \ du$

2. Originally Posted by Air
$\int \frac{u}{\sqrt{u^2 +16}} \ du$

Make the substitution $w = u^2 + 16$. Note that $\frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u}$ ....

3. Hi

$\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?$

4. Originally Posted by flyingsquirrel
Hi

$\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?$
$\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right) = \frac{ \mathrm{d}}{\mathrm{d}u} \left( u^2+16 \right)^\frac{1}{2} = \frac{1}{2}(u^2 + 16)^{-\frac{1}{2}}(2u) = (u)(u^2 + 16)^{-\frac{1}{2}} = \frac{u}{\sqrt{u^2+16}}$

5. Originally Posted by mr fantastic
Make the substitution $w = u^2 + 16$. Note that $\frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u}$ ....
You should try this way to see why flyingsquirrel was able to suggest you consider the derivative $\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)$ ......

6. Originally Posted by mr fantastic
You should try this way to see why flyingsquirrel was able to suggest you consider the derivative $\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)$ ......
$\int \frac{u}{\sqrt{u^2 + 16}} \ du$

$w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$

$\int \frac{u}{\sqrt w} . \frac{dw}{2u}$

$\int \frac{1}{\sqrt w} . \frac{dw}{2}$

$\frac{1}{2} \int \frac{1}{\sqrt w} \ dw$

$\frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right]$

$w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$

7. Originally Posted by Air
$\int \frac{u}{\sqrt{u^2 + 16}} \ du$

$w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$

$\int \frac{u}{\sqrt w} . \frac{dw}{2u}$

$\int \frac{1}{\sqrt w} . \frac{dw}{2}$

$\frac{1}{2} \int \frac{1}{\sqrt w} \ dw$

$\frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right]$

$w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$
+ C.