# Thread: Find the value of h given the following relation?

1. ## Find the value of h given the following relation?

If:
∫(vo to 0) (v/(-g - (kv^2/m) dv = ∫(0 to h) dh
, where k and g are constants,

Find h.

(What is the integral of type ∫ (x/((ab - cx^2)/a)) dx, where a, b and c are constants?)

2. Originally Posted by fardeen_gen
If:
(What is the integral of type ∫ (x/((ab - cx^2)/a)) dx, where a, b and c are constants?)
$\int {\frac{x}{{\left( {{{\left( {ab - c{x^2}} \right)} \mathord{\left/{\vphantom {{\left( {ab - c{x^2}} \right)} a}} \right.\kern-\nulldelimiterspace} a}} \right)}}dx} = \frac{a}{{2c}}\int {\frac{{2cx}}{{ab - c{x^2}}}dx} =$ $- \frac{a}{{2c}}\int {\frac{{d\left( {ab - c{x^2}} \right)}}{{ab - c{x^2}}}} = - \frac{a}{{2c}}\ln \left| {ab - c{x^2}} \right| + C.$

3. Originally Posted by DeMath
$\int {\frac{x}{{\left( {{{\left( {ab - c{x^2}} \right)} \mathord{\left/{\vphantom {{\left( {ab - c{x^2}} \right)} a}} \right.\kern-\nulldelimiterspace} a}} \right)}}dx} = \frac{a}{{2c}}\int {\frac{{2cx}}{{ab - c{x^2}}}dx} =$ $- \frac{a}{{2c}}\int {\frac{{d\left( {ab - c{x^2}} \right)}}{{ab - c{x^2}}}} = - \frac{a}{{2c}}\ln \left| {ab - c{x^2}} \right| + C.$
Is -2cx dx same as d(ab -cx^2)?How?

4. Since: $\frac{d}{dx} \left(ab - cx^2\right) = -2cx$

In the standard u-sub notation, it is just recognizing that: ${\color{blue}u = ab - cx^2} \ \Rightarrow \ du = -2cx \ dx \ \Leftrightarrow \ {\color{red}-\frac{du}{2c} = x dx}$

So: $a \int \frac{{\color{red}x}}{{\color{blue}ab-x^2}} \ {\color{red}dx} \ = \ a \int \frac{{\color{red}-\frac{du}{2c}}}{{\color{blue}u}} \ = \ -\frac{a}{2c} \int \frac{du}{u} = \ \cdots$