integration suitable substituition

Apr 2014
275
1
canada
for this question, the question only stated SUITABLE substituition, what substituition should i use? this substituion does not involve trigo functions , am i right?DSC_0143~2[1].jpg
 

Prove It

MHF Helper
Aug 2008
12,894
5,000
I'd write $\displaystyle \begin{align*} \int_0^4{\frac{1}{\left( 1 + \sqrt{x} \right) ^2}\,dx} &= \int_0^4{\frac{2\sqrt{x}}{2\sqrt{x}\left( 1 + \sqrt{x} \right) ^2}\,dx} \\ &= 2\int_0^4{ \frac{\sqrt{x}}{\left( 1 + \sqrt{x} \right) ^2} \left( \frac{1}{2\sqrt{x}} \right) \, dx } \end{align*}$

So now make the substitution $\displaystyle \begin{align*} u = 1 + \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(4) = 3 \end{align*}$, the integral becomes

$\displaystyle \begin{align*} 2\int_0^4{\frac{\sqrt{x}}{\left( 1 + \sqrt{x} \right) ^2} \left( \frac{1}{2\sqrt{x}} \right) \, dx} &= 2\int_1^3{\frac{u - 1}{u^2}\,du} \\ &= 2\int_1^3{ u^{-1} - u^{-2}\,du } \\ &= 2 \left[ \ln{|u|} + u^{-1} \right] _1^3 \\ &= 2\left[ \left( \ln{|3|} + 3^{-1} \right) - \left( \ln{|1|} + 1^{-1} \right) \right] \\ &= 2 \left[ \ln{(3)} + \frac{1}{3} - 0 - 1 \right] \\ &= 2\left[ \ln{(3)} - \frac{2}{3} \right] \end{align*}$