1. ## hard integral(i think)

integrate (4) / (4(x^1/2) + 4)

i believe i need a u substitution, but i am not sure

2. Originally Posted by whocares27
integrate (4) / (4(x^1/2) + 4)

i believe i need a u substitution, but i am not sure
Make the substitution $\displaystyle u = \sqrt{x} \Rightarrow dx = 2u \, du$.

3. Originally Posted by mr fantastic
Make the substitution $\displaystyle u = \sqrt{x} \Rightarrow dx = 2u \, du$.
After which use some simple algebra to get $\displaystyle 2 \int 1 - \frac{1}{u+1} \, du$.

4. $\displaystyle \begin{gathered} \int {\frac{4} {{4\sqrt x + 4}}dx} \hfill \\ \int {\frac{{dx}} {{\sqrt x + 1}}} \hfill \\ u = \sqrt x \hfill \\ du = \tfrac{{dx}} {{2\sqrt x }} = \tfrac{{dx}} {{2u}} \Rightarrow dx = 2udu \hfill \\ \end{gathered}$

$\displaystyle 2\int {\frac{{udu}} {{u + 1}}} = 2\int {\frac{{u + 1 - 1}} {{u + 1}}du = 2\int {1 - \frac{1} {{u + 1}}du = 2u - 2\ln |u + 1| + C} }$

Then substitute the square root of x back in for u and your set.

5. thank you both so very much