1. ## URGENT help on intergrals pleaseee

Tan -1x
1 + x(squared) dx

and.

4cos3s ds

many thanks people

2. Originally Posted by Stockport_ben

Tan -1x
1 + x(squared) dx

and.

4cos3s ds

many thanks people
i suppose the first question is this..

$\displaystyle \int \frac{\tan ^{-1} x}{1+x^2} \, dx$

set $\displaystyle u = \tan ^{-1} x \Rightarrow du = \frac{1}{1+x^2} \, dx$ and that should be fine..

for the second one, set $\displaystyle u = 3s \Rightarrow du = 3ds$ and you can do it all the way..

3. heya yea thst right for the first one could you please do a step by step as im relativeley know to all this intergral stuff if not dont worry
thanks for the help

4. ok..

$\displaystyle \int \frac{\tan ^{-1} x}{1+x^2} \, dx$

set $\displaystyle u = \tan ^{-1} x \Rightarrow du = \frac{1}{1+x^2} \, dx$

then $\displaystyle \int \frac{\tan ^{-1} x}{1+x^2} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\tan^{-1} x)^2}{2} + C$..

is that ok?

5. yea thats perfect thanks