Integration help urgent C4

mcfc2009

Can someone explain to me why the integral of 1/root(1+2x) = root(1+2x)

thanks

Original post by mcfc2009

Can someone explain to me why the integral of 1/root(1+2x) = root(1+2x)

thanks

\displaystyle \int \frac1{\sqrt{1+2x}} \ dx = \int (1+2x)^{ - \frac12} \ dx

Now its simple integration.

Reply 2

TenOfThem

What do you get when you differentiate

(1+2x)^{\frac{1}{2}}

2(1+2x)^1/2

Original post by TenOfThem

What do you get when you differentiate

(1+2x)^{\frac{1}{2}}

I get 2(1+2x)^1/2

Reply 5

TenOfThem

Original post by mcfc2009

I get 2(1+2x)^1/2

Why haven't you multiplied by and decreased the power

Reply 6

mcfc2009

Original post by TenOfThem

Why haven't you multiplied by and decreased the power

Ive brought it up to the top and made it (1+2x)^-1/2 increase the power by 1 to 1/2 then divided by a half which then gave me 2(1+2x)^1/2

Reply 7

TenOfThem

Original post by mcfc2009

Ive brought it up to the top and made it (1+2x)^-1/2 increase the power by 1 to 1/2 then divided by a half which then gave me 2(1+2x)^1/2

Sorry

You have lost me

I thought you were answering the question that I asked and you quoted

Reply 8

OSharp

+c

I hate integration so much.

Reply 9

raheem94

Original post by mcfc2009

Ive brought it up to the top and made it (1+2x)^-1/2 increase the power by 1 to 1/2 then divided by a half which then gave me 2(1+2x)^1/2

\displaystyle \int (ax+b)^{ d } \ dx = \frac1{d+1} (ax+b)^{d+1} \times \frac1{a} +C

Reply 10

mcfc2009

Original post by TenOfThem

Sorry

You have lost me

I thought you were answering the question that I asked and you quoted

it started as 1/(1+2x)^1/2 then I followed the method i stated above.

Reply 11

TenOfThem

Original post by mcfc2009

it started as 1/(1+2x)^1/2 then I followed the method i stated above.

Yes, I realise that ... but, as you know you got it wrong ...

I was asking you what you get when you differentiate

(1+2x)^{\frac{1}{2}}

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