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 $u=2x-1$ so $du=2dx$

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

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

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

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

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

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

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

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

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

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 !! =)
$\frac{dt}{dx} = \frac{1}{2(4 - x)^2}$

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

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

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

So

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

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

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

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

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

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

Now use your initial condition to find C.