# Thread: Integration by substitution - where am I going wrong?

1. ## Integration by substitution - where am I going wrong?

The integral is:

The substitution is: u = sqrt[x]

I don't know where I am going wrong, I am wondering if hte text book i'm using could have a typo or something (these books do occasionally get stuff wrong).

So what I did:

u=sqrt[x] => x=u^2

du/dx = (x^-1/2)/2 => dx = 2udu

subtituting into the original integral and multiplying out we now have:

And now i have no idea what to do :/

According to the text book, the final answer is:

1/2*ln(9/5)

Where am I going wrong?

2. In the integral $\displaystyle \int_1^2\frac{2u}{4u^3-u}\,du$, divide top and bottom of the fraction by u, then use partial fractions.

3. Originally Posted by GRM92
The integral is:

The substitution is: u = sqrt[x]

I don't know where I am going wrong, I am wondering if hte text book i'm using could have a typo or something (these books do occasionally get stuff wrong).

So what I did:

u=sqrt[x] => x=u^2

du/dx = (x^-1/2)/2 => dx = 2udu

subtituting into the original integral and multiplying out we now have:

And now i have no idea what to do :/ Mr F says: Now use a partial fraction decomposition.

According to the text book, the final answer is:

1/2*ln(9/5)

Where am I going wrong?
..