Question

\[\int\limits {1 \over 3 \sqrt x +4x}dx--->x=u^2\]\[\int\limits {1 \over 3u + 4u^2}du\]How to carry on?

Answer 1

Can you show the steps?

Answer 2

u have missed 2udu iin the numerator
and then u from denom and num cancels out

and then u can assume 3+4u=z (or anything u wish)
and then proceed to get the result
the answer is i think (1/2)ln(3+4√x) +c

Answer 3

I don't know how to integrate this... can you help and show step by step?