# Intergration by substitution

• thomas49th
In summary: Right let's rearrange u in terms of x (i think that's how you say it):x = \frac{u^{2} - 1}{2}And now get an expression for dxu = (2x-1)^{\frac{1}{2}}use the chain rule on it to give\frac{dx}{du} (2x-1)^{-\frac{1}{2}}= dx = \frac{du}{(2x-1)^{-\frac{1}{2}}}Right now substitute that in \int^{5}_{1} \frac{3\frac{u^{

## Homework Statement

Using the substitution u² = 2x - 1, or otherwise, find the exact value of
$$\int^{5}_{1} \frac{3x}{\sqrt{2x-1}}dx$$

## The Attempt at a Solution

Right let's rearrange u in terms of x (i think that's how you say it):

$$x = \frac{u^{2} - 1}{2}$$

And now get an expression for dx

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

use the chain rule on it to give

$$\frac{dx}{du} (2x-1)^{-\frac{1}{2}}$$

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

Right now substitute that in

$$\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}$$

now according to the mark scheme

$$\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}$$

can be simplified

$$\int^{5}_{1} \frac{3u^{2} - 3}{2u}\frac{du}{(2x-1)^{-\frac{1}{2}}}$$

shouldn't the bit here:

$$\frac{3u^{2} - 3}{2u}$$

be

$$\frac{3u^{2} - 3}{6u}$$

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thomas49th said:
Right let's rearrange u in terms of x (i think that's how you say it):

$$x = \frac{u^{2} - 1}{2}$$

Should be...

$$x = \frac{u^{2} + 1}{2}$$

Shouldn't it?

I managed to get the integral with 2u on the bottom and looking the same, but let me know how you're making out.

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woops my bad i did 3 x EVERYTHING On the fraction (including the denominator). didn;t really have my brain in gear

sorry

cheers tho :)

Haha no worries, least you found out where you went wrong

thomas49th said:

## Homework Statement

Using the substitution u² = 2x - 1, or otherwise, find the exact value of
$$\int^{5}_{1} \frac{3x}{\sqrt{2x-1}}dx$$

## The Attempt at a Solution

Right let's rearrange u in terms of x (i think that's how you say it):

$$x = \frac{u^{2} - 1}{2}$$

And now get an expression for dx

$$u = (2x-1)^{\frac{1}{2}}$$
It's much simpler not to rearrange u in terms of x (or "solve for x"). Instead, from u2= 2x- 1, 2udu= 2dx so udu= dx.

use the chain rule on it to give

$$\frac{dx}{du} (2x-1)^{-\frac{1}{2}}$$

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

Right now substitute that in

$$\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}$$

now according to the mark scheme

$$\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}$$

can be simplified

$$\int^{5}_{1} \frac{3u^{2} - 3}{2u}\frac{du}{(2x-1)^{-\frac{1}{2}}}$$

shouldn't the bit here:

$$\frac{3u^{2} - 3}{2u}$$

be

$$\frac{3u^{2} - 3}{6u}$$