1. integration problems

could someone help me integrate the following equations because i keep getting the wrong answer and i dont no why T.T thankyou !

a) x/[(2x-1)^1/2]
b) dt/dx = 1/{2[(4-x)^2]} and that x = 0 when t=0 find x in terms of t

btw i am using the u substitution method for these two questions

thankyou so much for any help given !! =)

2. Let $\displaystyle u=2x-1$ so $\displaystyle du=2dx$

So $\displaystyle \int {xdx\over \sqrt{2x-1}} ={1\over 4} \int {(2x)(2dx)\over \sqrt{2x-1}}$

$\displaystyle ={1\over 4} \int {(u+1)du\over \sqrt{u}} ={1\over 4} \int (u^{1/2}+u^{-1/2})du$

3. Originally Posted by flyinhigh123
could someone help me integrate the following equations because i keep getting the wrong answer and i dont no why T.T thankyou !

a) x/[(2x-1)^1/2]
b) dt/dx = 1/{2[(4-x)^2]} and that x = 0 when t=0 find x in terms of t

btw i am using the u substitution method for these two questions

thankyou so much for any help given !! =)
For a), you can use integration by parts.

Remember that $\displaystyle \int{u\,dv} = uv - \int{v\,du}$.

$\displaystyle \int{\frac{x}{(2x - 1)^{\frac{1}{2}}}\,dx} = \int{x(2x - 1)^{-\frac{1}{2}}\,dx}$.

Let $\displaystyle u = x$ so that $\displaystyle du = 1$.

Let $\displaystyle dv = (2x - 1)^{-\frac{1}{2}}$ so that $\displaystyle v = (2x - 1)^{\frac{1}{2}}$.

So $\displaystyle \int{\frac{x}{(2x - 1)^{\frac{1}{2}}}\,dx} = \int{x(2x - 1)^{-\frac{1}{2}}\,dx}$

$\displaystyle = x(2x - 1)^{\frac{1}{2}} - \int{(2x - 1)^{\frac{1}{2}}\,dx}$

$\displaystyle = x(2x - 1)^{\frac{1}{2}} - \frac{1}{3}(2x - 1)^{\frac{3}{2}} + C$.

4. Originally Posted by flyinhigh123
could someone help me integrate the following equations because i keep getting the wrong answer and i dont no why T.T thankyou !

a) x/[(2x-1)^1/2]
b) dt/dx = 1/{2[(4-x)^2]} and that x = 0 when t=0 find x in terms of t

btw i am using the u substitution method for these two questions

thankyou so much for any help given !! =)
$\displaystyle \frac{dt}{dx} = \frac{1}{2(4 - x)^2}$

$\displaystyle = \frac{1}{2}(4 - x)^{-2}$

$\displaystyle = -\frac{1}{2}(4 - x)^{-2}(-1)$.

Let $\displaystyle u = 4 - x$ so that $\displaystyle \frac{du}{dx} = -1$

So

$\displaystyle t = \int{ \frac{1}{2(4 - x)^2}\,dx}$

$\displaystyle = -\frac{1}{2}\int{(4 - x)^{-2}(-1)\,dx}$

$\displaystyle = -\frac{1}{2}\int{u^{-2}\,\frac{du}{dx}\,dx}$

$\displaystyle = -\frac{1}{2}\int{u^{-2}\,du}$

$\displaystyle = \frac{1}{2}u^{-1} + C$

$\displaystyle = \frac{1}{2(4 - x)} + C$.

Now use your initial condition to find C.