# Can't figure out how to find the integral of this equation

• Jul 21st 2009, 02:32 PM
boiler1987
Can't figure out how to find the integral of this equation
how would I solve the integral of 1/(a+b*sqrt(y))dy

i've tried various approaches but I can't seem to get anywhere. Anyone have any strategies or solutions? (Also checked quite a few integral tables, can't find it anywhere)

also the value at y=0 should be 0.
• Jul 21st 2009, 02:47 PM
Plato
Go to this website W|A
In the input window, type in this exact phrase: integrate 1/[a +b sqrt[y]]dy.
Click the equals bar at the end if the input window.
Be sure you click ‘show steps’ to see the solution.
• Jul 21st 2009, 02:52 PM
DeMath
Quote:

Originally Posted by boiler1987
how would I solve the integral of 1/(a+b*sqrt(y))dy

i've tried various approaches but I can't seem to get anywhere. Anyone have any strategies or solutions? (Also checked quite a few integral tables, can't find it anywhere)

also the value at y=0 should be 0.

$\int {\frac{{dy}}{{a + b\sqrt y }}} = \left\{ \begin{gathered}y = {t^2}, \hfill \\dy = 2tdt \hfill \\ \end{gathered} \right\} = 2\int {\frac{t}{{a + bt}}} dt =$

$= \frac{2}{b}\int {\frac{{a + bt - a}}{{a + bt}}} dt = \frac{2}{b}\int {\left( {1 - \frac{a}{{a + bt}}} \right)} dt =$

$= \frac{2}{b}t - \frac{{2a}}{b}\int {\frac{{dt}}{{a + bt}}} = \frac{2}{b}t - \frac{{2a}}{{{b^2}}}\int {\frac{{d\left( {a + bt} \right)}}{{a + bt}}} =$

$= \frac{2}{b}t - \frac{{2a}}{{{b^2}}}\ln \left| {a + bt} \right| + C = \frac{2}{b}\sqrt y - \frac{{2a}}{{{b^2}}}\ln \left| {a + b\sqrt y } \right| + C.$